Numbers

1. Core Concepts & Theoretical Blueprint

The Number System chapter classifies numbers into structural categories and provides tools — divisibility rules, the division algorithm, unit-digit cyclicity, and factor-counting formulas — to answer structural questions about any integer without full computation.

Classification of Numbers:

• Natural Numbers (N): 1, 2, 3, 4, ... (positive counting numbers)
• Whole Numbers (W): 0, 1, 2, 3, ... (natural numbers + zero)
• Integers (Z): ..., −2, −1, 0, 1, 2, ...
• Rational Numbers: expressible as $\frac{p}{q}$, $q\neq0$, $p,q$ integers
• Irrational Numbers: non-terminating, non-repeating decimals (e.g., $\sqrt2, \pi$)
• Prime Numbers: exactly two distinct positive factors (1 and itself); smallest prime is 2 (the only even prime)
• Composite Numbers: more than two factors
• Note: 1 is neither prime nor composite.

Division Algorithm (the single most reused identity in this chapter):

Divisibility Rules Table (must be instantly recallable):

Divisor	Rule
2	Last digit is even (0,2,4,6,8)
3	Sum of digits divisible by 3
4	Last two digits divisible by 4
5	Last digit is 0 or 5
6	Divisible by both 2 and 3
7	Double the last digit, subtract from remaining number; result divisible by 7 (iterative)
8	Last three digits divisible by 8
9	Sum of digits divisible by 9
10	Last digit is 0
11	(Sum of digits at odd places) − (Sum of digits at even places) divisible by 11 (including 0)
12	Divisible by both 3 and 4
25	Last two digits divisible by 25 (00, 25, 50, 75)

Unit Digit Cyclicity Table (powers cycle with period ≤4):

Base unit digit	Cycle of unit digits	Cycle length
0,1,5,6	always 0,1,5,6 respectively	1
4	4, 6	2
9	9, 1	2
2	2,4,8,6	4
3	3,9,7,1	4
7	7,9,3,1	4
8	8,4,2,6	4

Number of Factors and Sum of Factors (from prime factorization $N=p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}$):

Summation Formulas (Series of Natural Numbers):

The Universal Trap: Three high-frequency traps:

1. Applying the divisibility-by-7/13 rule incorrectly or too slowly under time pressure — for large numbers, direct division is often faster than the iterative rule; know when to skip the "rule" and just divide.
2. Miscounting the unit-digit cycle position — students often forget to first reduce the EXPONENT using the cycle length (via $\text{exponent} \bmod \text{cycle length}$, treating a 0 remainder as the LAST position in the cycle, not position 0) before reading off the unit digit.
3. Confusing "number of factors" with "number of PRIME factors" — a question asking "how many factors does 360 have" wants the full count via $(a_1+1)(a_2+1)\cdots$, not just the count of distinct primes in its factorization.

2. Exhaustive Question Typology

                                NUMBERS
                                   |
    -----------------------------------------------------------------------
    |             |               |               |               |       |
Type 1:        Type 2:         Type 3:         Type 4:         Type 5:  Type 6:
Divisibility   Division        Unit Digit      Number of       Sum of   Sum of
Rule           Algorithm       of Powers       Factors/         Factors  Natural/
Application    (Dividend =                     Divisors of a            Square/Cube
               Divisor×Q+R)                    Number                   Series
    |             |               |
Type 7:        Type 8:         Type 9:
Combined       Face Value &    Remainder-
Divisibility   Place Value     based
(divisible by                 Problems
composite                     (using mod
numbers, e.g.                 arithmetic,
88=8×11)                      not full
                               division)

Type 1 — Divisibility rule application:

• Core Scenario: "Is the number N divisible by d?" or "Find the digit x such that the number is divisible by d."
• Governing Equation: Apply the specific rule for d from Section 1's table; for unknown-digit problems, set up the rule as an equation/condition on x and solve.

Type 2 — Division algorithm problems:

• Core Scenario: "A number when divided by d gives quotient q and remainder r. Find the number," or reverse (find remainder/quotient given the number).
• Governing Equation: $N=dq+r$

Type 3 — Unit digit of powers:

• Core Scenario: "Find the unit digit of $17^{83}$" or "find the unit digit of a product/sum of several powers."
• Governing Equation: Reduce exponent using cycle length: find e' = \text{exponent}\bmod(\text{cycle length}) (using cycle length as the value when remainder is 0), then read off the e'-th value in the base's cycle table.

Type 4 — Number of factors/divisors of a number:

• Core Scenario: "Find the total number of factors of 1800."
• Governing Equation: Prime factorize $N=p_1^{a_1}p_2^{a_2}\cdots$, then Number of factors $=(a_1+1)(a_2+1)\cdots$

Type 5 — Sum of factors of a number:

• Core Scenario: "Find the sum of all factors of 360."
• Governing Equation: $\text{Sum} = \prod_i \dfrac{p_i^{a_i+1}-1}{p_i-1}$

Type 6 — Sum of natural number/square/cube series:

• Core Scenario: "Find the sum of the first 30 natural numbers," or "sum of squares/cubes of the first n numbers."
• Governing Equation: Use the appropriate Section 1 summation formula directly.

Type 7 — Combined divisibility (divisible by a composite number via its coprime factors):

• Core Scenario: "Is N divisible by 88?" (recognize 88 = 8 × 11, both coprime factors, and check divisibility by each independently).
• Governing Equation: N divisible by $m\times n$ (m, n coprime) $\iff$ N divisible by m AND N divisible by n.

Type 8 — Face value and place value problems:

• Core Scenario: "In the number 84,532, find the difference between the place value and face value of the digit 5."
• Governing Equation: Place value $=$ digit $\times$ (power of 10 based on position); Face value $=$ the digit itself, unconditionally.

Type 9 — Remainder-based problems using modular reasoning (without full division):

• Core Scenario: "Find the remainder when $2^{100}$ is divided by 7," or similar large-exponent/large-number remainder problems.
• Governing Equation: Use cyclicity of remainders: find the pattern of $(\text{base}^k \bmod \text{divisor})$ for small k, identify the cycle length, then reduce the exponent modulo the cycle length exactly as in Type 3's unit-digit method, but tracking full remainders instead of just unit digits.

3. Type-wise Practice MCQs with Full Solutions

Type 1 — Divisibility Rule Application

MCQ 1. Which of the following numbers is divisible by 11? (A) 25784 (B) 39284 (C) 47531 (D) 52938

Correct Answer: (A) Solution: For 25784: odd-place digits (from right) are 4,7,5 → sum=16; even-place digits are 8,2 → sum=10. Wait, recompute carefully with standard right-to-left alternating sum: digits are 2-5-7-8-4. From right: position1=4,pos2=8,pos3=7,pos4=5,pos5=2. Sum of odd positions (1,3,5) = 4+7+2=13. Sum of even positions (2,4)=8+5=13. Difference=0, which is divisible by 11. 25784 is divisible by 11.

MCQ 2. Find the smallest digit to be placed in the blank so that the number 3_45 is divisible by 3. (A) 0 (B) 1 (C) 2 (D) 3

Correct Answer: (A) Solution: Digit sum without blank $=3+4+5=12$, already divisible by 3. The smallest digit that keeps the sum divisible by 3 is 0 (since 12+0=12 is divisible by 3).

MCQ 3. What least value should be given to * so that the number 653*47 is divisible by 11? (A) 5 (B) 4 (C) 3 (D) 6

Correct Answer: (A) Solution: Digits (left to right): 6,5,3,,4,7. From the right: pos1=7,pos2=4,pos3=,pos4=3,pos5=5,pos6=6. Odd positions (1,3,5): 7++5=12+. Even positions (2,4,6): 4+3+6=13. Difference $=(12+<em>)-13=</em>-1$, which must be 0 or a multiple of 11. Smallest non-negative: $*-1=0\Rightarrow *=1$... Let's re-verify against option set; if *=1 works, adjust options to reflect verified answer.

MCQ 3 (verified). What least value should be given to * so that the number 653*47 is divisible by 11? (A) 1 (B) 4 (C) 3 (D) 6

Correct Answer: (A) Solution: As derived, $*-1=0\Rightarrow *=1$ is the least valid digit satisfying divisibility by 11.

Type 2 — Division Algorithm

MCQ 1. A number when divided by 342 gives a remainder of 47. Find the remainder when the same number is divided by 18. (A) 11 (B) 9 (C) 5 (D) 13

Correct Answer: (A) Solution: Since $342=18\times19$, the number $N=342k+47$. Dividing by 18: $342k$ is exactly divisible by 18 (since 342 is a multiple of 18), so the remainder comes purely from 47 divided by 18: $47=18\times2+11$. Remainder $=11$.

MCQ 2. On dividing a number by 68, the quotient is 269 and the remainder is 27. Find the number. (A) 18319 (B) 18292 (C) 18300 (D) 18350

Correct Answer: (A) Solution: $N=d\times q+r=68\times269+27=18292+27=18319$.

MCQ 3. A number, when divided by 899, gives a remainder 63. Find the remainder when the same number is divided by 29. (A) 5 (B) 4 (C) 6 (D) 3

Correct Answer: (A) Solution: $899=29\times31$, so N is a multiple of 29 plus 63. $63=29\times2+5$. Remainder $=5$.

Type 3 — Unit Digit of Powers

MCQ 1. Find the unit digit of $7^{45}$. (A) 7 (B) 9 (C) 3 (D) 1

Correct Answer: (A) Solution: Cycle for base ending in 7: [7,9,3,1], length 4. $45\bmod4=1$, so use position 1 in cycle → unit digit = 7.

MCQ 2. Find the unit digit of $24^{24}$. (A) 6 (B) 4 (C) 2 (D) 8

Correct Answer: (A) Solution: Base unit digit = 4, cycle [4,6], length 2. $24\bmod2=0$, use last (2nd) position in cycle → unit digit = 6.

MCQ 3. Find the unit digit of $3^{57}\times2^{35}$. (A) 4 (B) 6 (C) 2 (D) 8

Correct Answer: (A) Solution: For $3^{57}$: cycle [3,9,7,1], length 4; $57\bmod4=1$→ unit digit 3. For $2^{35}$: cycle [2,4,8,6], length 4; $35\bmod4=3$→ unit digit 8. Product's unit digit = unit digit of $3\times8=24$ → 4.

Type 4 — Number of Factors/Divisors

MCQ 1. Find the total number of factors of 360. (A) 24 (B) 20 (C) 18 (D) 16

Correct Answer: (A) Solution: $360=2^3\times3^2\times5^1$. Number of factors $=(3+1)(2+1)(1+1)=4\times3\times2=24$.

MCQ 2. How many factors does 1024 have? (A) 11 (B) 10 (C) 9 (D) 12

Correct Answer: (A) Solution: $1024=2^{10}$. Number of factors $=(10+1)=11$.

MCQ 3. Find the number of even factors of 180. (A) 12 (B) 18 (C) 9 (D) 6

Correct Answer: (A) Solution: $180=2^2\times3^2\times5^1$. Total factors $=(2+1)(2+1)(1+1)=18$. Odd factors (ignoring the 2's, i.e. factors of $3^2\times5^1$) $=(2+1)(1+1)=6$. Even factors $=$ Total $-$ Odd $=18-6=12$.

Type 5 — Sum of Factors

MCQ 1. Find the sum of all factors of 60. (A) 168 (B) 150 (C) 180 (D) 200

Correct Answer: (A) Solution: $60=2^2\times3^1\times5^1$. Sum $=\dfrac{2^3-1}{2-1}\times\dfrac{3^2-1}{3-1}\times\dfrac{5^2-1}{5-1}=7\times4\times6=168$.

MCQ 2. Find the sum of all factors of 100. (A) 217 (B) 200 (C) 210 (D) 225

Correct Answer: (A) Solution: $100=2^2\times5^2$. Sum $=\dfrac{2^3-1}{1}\times\dfrac{5^3-1}{4}=7\times\dfrac{124}{4}=7\times31=217$.

MCQ 3. Find the sum of all the odd factors of 180. (A) 24 (B) 39 (C) 168 (D) 91

Correct Answer: (B) Solution: $180=2^2\times3^2\times5^1$. Odd factors come from $3^2\times5^1$ only (ignore the powers of 2). Sum of factors of $3^2\times5$: $\dfrac{3^3-1}{2}\times\dfrac{5^2-1}{4}=\dfrac{26}{2}\times\dfrac{24}{4}=13\times6=78$. (Recheck: this gives 78, not matching option B; correcting the option set.)

MCQ 3 (verified). Find the sum of all the odd factors of 180. (A) 78 (B) 39 (C) 168 (D) 91

Correct Answer: (A) Solution: As derived, sum of odd factors (i.e., all factors of $3^2\times5^1=45$) $=\dfrac{3^3-1}{3-1}\times\dfrac{5^2-1}{5-1}=13\times6=78$.

Type 6 — Sum of Natural/Square/Cube Series

MCQ 1. Find the sum of the first 40 natural numbers. (A) 820 (B) 800 (C) 780 (D) 840

Correct Answer: (A) Solution: $\dfrac{n(n+1)}{2}=\dfrac{40\times41}{2}=20\times41=820$.

MCQ 2. Find the sum of squares of the first 10 natural numbers. (A) 385 (B) 350 (C) 400 (D) 330

Correct Answer: (A) Solution: $\dfrac{n(n+1)(2n+1)}{6}=\dfrac{10\times11\times21}{6}=\dfrac{2310}{6}=385$.

MCQ 3. Find the sum of cubes of the first 6 natural numbers. (A) 441 (B) 400 (C) 360 (D) 484

Correct Answer: (A) Solution: $\left[\dfrac{n(n+1)}{2}\right]^2=\left[\dfrac{6\times7}{2}\right]^2=21^2=441$.

Type 7 — Combined Divisibility

MCQ 1. Is 4356 divisible by 12? (A) Yes (B) No (C) Only by 6, not 12 (D) Cannot be determined

Correct Answer: (A) Solution: $12=4\times3$ (coprime factors). Last two digits "56" → $56\div4=14$, divisible by 4. Digit sum $=4+3+5+6=18$, divisible by 3. Since divisible by both 4 and 3, it is divisible by 12.

MCQ 2. Which of the following is divisible by 88? (A) 3872 (B) 4562 (C) 5236 (D) 6314

Correct Answer: (A) Solution: $88=8\times11$. Check 3872: last three digits 872 ÷ 8 = 109, divisible by 8. Alternating sum (from right): pos1=2,pos2=7,pos3=8,pos4=3; odd positions(1,3)=2+8=10, even positions(2,4)=7+3=10, difference=0, divisible by 11. Since divisible by both 8 and 11, 3872 is divisible by 88.

MCQ 3. Find the least value of x such that 653x is divisible by both 8 and 9. (A) No such single digit x exists such that both conditions hold simultaneously for 653x — verify each option digit for consistency. (B)-(D) placeholders

(Note: to keep this MCQ standard-exam calibrated, restate with a genuinely 4-digit form allowing both conditions.)

MCQ 3 (restated, standard form). Find the value of x such that the number 24x38 is divisible by 8 (Type 7 focus: isolating just the "last three digits divisible by 8" sub-rule as one coprime-factor check within a combined-divisibility context). (A) x can be any digit since only the last 3 digits (x38) matter; find x such that x38 is divisible by 8 — check x=2: 238÷8=29.75 (no); x=6: 638÷8=79.75(no); x=1:138÷8=17.25(no); test systematically to find x=7: 738÷8=92.25(no); the correct approach is x38 divisible by 8 requires solving directly. (B) x = 3 (C) x = 5 (D) x = 9

Correct Answer: Let us verify directly: 038 mod 8 = 38 mod 8 = 6 (x=0); 138 mod 8=138-136=2(x=1); 238mod8=238-232=6(x=2); 338mod8=338-336=2(x=3); 438mod8=438-432=6(x=4); 538mod8=538-536=2(x=5); 638mod8=638-632=6(x=6); 738mod8=738-736=2(x=7); 838mod8=838-832=6(x=8); 938mod8=938-936=2(x=9). None give remainder 0 — this reveals x38 is never divisible by 8 for any digit x, since 38 mod 8=6 is fixed by the last two digits' contribution pattern here; recognize this MCQ instead as a checking exercise. For exam purposes, replace with a cleaner instance:

MCQ 3 (final clean version). Find the digit x such that the number 24x36 is divisible by 8. (A) x = 1 (B) x = 3 (C) x = 5 (D) x = 7

Correct Answer: (A) Solution: Only the last three digits "x36" matter for divisibility by 8. Test x=1: 136÷8=17, exact. So x=1 works, confirming (A).

Type 8 — Face Value and Place Value

MCQ 1. In the number 57,942, find the place value of the digit 7. (A) 7000 (B) 700 (C) 70 (D) 7

Correct Answer: (A) Solution: The digit 7 is in the thousands place (5-7-9-4-2), so its place value $=7\times1000=7000$.

MCQ 2. In the number 8,53,192, find the difference between the place value and the face value of the digit 5. (A) 49,995 (B) 50,000 (C) 45,000 (D) 49,500

Correct Answer: (A) Solution: The digit 5 is in the ten-thousands place: place value $=5\times10000=50000$. Face value $=5$. Difference $=50000-5=49995$.

MCQ 3. Find the sum of the place values of both occurrences of 4 in the number 42,347. (A) 40,004 (B) 44,000 (C) 8000 (D) 40,400

Correct Answer: (A) Solution: First 4 (ten-thousands place): $4\times10000=40000$. Second 4 (units place... check number 4-2-3-4-7): the second 4 is in the tens place: $4\times10=40$. Wait — position check: digits are 4(ten-thousands),2(thousands),3(hundreds),4(tens),7(units). So second 4's place value $=4\times10=40$. Sum $=40000+40=40040$. (Recheck against option; correcting option set to reflect verified value.)

MCQ 3 (verified). Find the sum of the place values of both occurrences of 4 in the number 42,347. (A) 40,040 (B) 44,000 (C) 8000 (D) 40,400

Correct Answer: (A) Solution: As derived: $40000+40=40040$.

Type 9 — Remainder-Based Problems (Modular Reasoning)

MCQ 1. Find the remainder when $2^{100}$ is divided by 7. (A) 2 (B) 4 (C) 1 (D) 6

Correct Answer: (B) Solution: Powers of 2 mod 7 cycle as: $2^1=2,2^2=4,2^3=1,2^4=2,...$ cycle length 3. $100\bmod3=1$, so remainder matches $2^1\bmod7=2$. (Recheck: cycle position 1 → remainder 2, not 4; correcting the marked answer.)

MCQ 1 (verified). Find the remainder when $2^{100}$ is divided by 7. (A) 2 (B) 4 (C) 1 (D) 6

Correct Answer: (A) Solution: As derived: cycle of $2^k\bmod7$ is [2,4,1] repeating with length 3. $100\bmod3=1$ (since $99=3\times33$), so remainder = 1st term in cycle = 2.

MCQ 2. Find the remainder when $3^{250}$ is divided by 5. (A) 4 (B) 1 (C) 3 (D) 2

Correct Answer: (A) Solution: Powers of 3 mod 5: $3^1=3,3^2=4,3^3=2,3^4=1$, cycle length 4. $250\bmod4=2$, so remainder = 2nd term = 4.

MCQ 3. Find the remainder when $17^{200}$ is divided by 18. (A) 1 (B) 17 (C) 16 (D) 2

Correct Answer: (A) Solution: $17\equiv-1\pmod{18}$. So $17^{200}\equiv(-1)^{200}=1\pmod{18}$ (since 200 is even). Remainder = 1.

4. High-Yield Speed Tricks & Shortcut Mental Models

Shortcut 1 — The "$-1$ Trick" for Remainders Near the Divisor

• Application: Any remainder problem where the base is exactly 1 less than or 1 more than the divisor (Type 9-style problems, like the $17^{200}\div18$ example).
• Mental Model: If base $\equiv -1\pmod{d}$ (i.e., base = d−1), then base$^n\equiv(-1)^n\pmod d$ — instantly 1 if n is even, or $d-1$ if n is odd. This converts a seemingly large power/remainder problem into a one-line parity check, without ever computing a cycle table.

Shortcut 2 — Coprime Factor Splitting for Composite-Divisor Divisibility

• Application: Any Type 7 "is N divisible by [composite number]" question (12, 18, 24, 88, 45, etc.).
• Mental Model: Always break the composite divisor into two coprime factors whose individual divisibility rules are simple (e.g., 12→3×4, 18→2×9, 88→8×11) and check each independently — never attempt a direct divisibility rule for the composite number itself, since none exists as cleanly as for its coprime components.

Shortcut 3 — Exponent Reduction via "mod cycle length, treat 0 as full cycle"

• Application: Every Type 3 (unit digit) and Type 9 (remainder) problem involving large exponents.
• Mental Model: Always compute (exponent mod cycle length) first. The one critical adjustment: if this comes out to exactly 0, do NOT read the "0th" position — instead use the LAST position of the cycle (since a remainder of 0 in modular exponent-reduction means the exponent is an exact multiple of the cycle, landing back on the final/complete cycle position, not a "zeroth" undefined one).

5. Deep-Dive: Most Frequently Asked Questions

Problem 1 (SSC/RRB Standard): Find the unit digit of the expression $3^{65}\times6^{59}\times7^{71}$.

Traditional Method (Slow): Compute unit digit of each term separately by writing out several powers manually to detect the pattern for each base (e.g., listing $3^1,3^2,3^3,3^4,3^5...$ until a repeat is visually spotted), then multiplying all three resulting digits together and taking the unit digit of that product. (This works but involves writing out multiple power sequences from scratch — ~45-50 seconds if the cycles aren't already memorized.)

Exam Shortcut (Fast): Recall cycles instantly: base 3 → [3,9,7,1], length 4; $65\bmod4=1$→ digit 3. Base 6 → always ends in 6 regardless of power → digit 6. Base 7 → [7,9,3,1], length 4; $71\bmod4=3$→ digit 3. Multiply unit digits: $3\times6\times3=54$→ unit digit 4. Answer: 4, obtained via three instant table lookups and one small multiplication — under 15 seconds, entirely bypassing any need to write out power sequences.

Problem 2 (UPSC/Banking Advanced): Find the number of zeros at the end of $100!$ (100 factorial).

Step-by-Step Breakdown:

1. Trailing zeros in a factorial are produced by factors of 10, i.e., pairs of (2×5) in the prime factorization of $100!$.
2. Since factors of 2 are far more abundant than factors of 5 in any factorial (every even number contributes a 2, but only every 5th number contributes a 5), the count of trailing zeros is determined entirely by the number of times 5 appears as a factor across all numbers from 1 to 100.
3. Use Legendre's formula for the highest power of a prime p dividing n!:
   $$\left\lfloor\frac{n}{p}\right\rfloor+\left\lfloor\frac{n}{p^2}\right\rfloor+\left\lfloor\frac{n}{p^3}\right\rfloor+\cdots$$
4. For n=100, p=5: $\left\lfloor\frac{100}{5}\right\rfloor=20$; $\left\lfloor\frac{100}{25}\right\rfloor=4$; $\left\lfloor\frac{100}{125}\right\rfloor=0$ (since 125>100, stop here).
5. Total power of 5 in $100!=20+4=24$.
6. Since powers of 2 in $100!$ vastly exceed 24, the limiting factor for forming pairs of (2×5) is exactly this count of 5's.
7. Answer: $100!$ has exactly 24 trailing zeros. The key structural insight (heavily tested across SSC/banking papers) is that trailing-zero-counting NEVER requires counting factors of 2 — only factors of 5, since 2's are never the bottleneck.

6. Chapter Checklist for Students

• I have the divisibility rules for 2 through 12 (and 25) memorized cold, with zero derivation time.
• I correctly reduce large exponents using (exponent mod cycle length), always mapping a 0 remainder to the FINAL position of the cycle, never a nonexistent "0th" position.
• I apply $(a_1+1)(a_2+1)\cdots$ directly for factor-counting and the standard geometric-series sum formula for sum-of-factors, from a verified prime factorization.
• I split composite-number divisibility checks (12, 18, 24, 88, etc.) into two coprime factors and verify each independently, rather than searching for a rule that doesn't exist.
• I know that trailing-zero-counting in factorials depends only on counting factors of 5 (via Legendre's formula), never factors of 2.