Study Guide · Chapter 30
14. Common Mistakes Aspirants Make
Free study material · concepts, shortcuts & solved questions
Select any text to highlight or save it
- Treating 1 as a prime number. 1 has only one factor (itself), so by definition it is neither prime nor composite. Correction: the smallest prime is 2.
- Assuming 0 is a natural number. In the SSC/RRB convention, Natural Numbers start from 1; Whole Numbers start from 0. Don’t mix the two sets up.
- Reversing the HCF × LCM = product relationship, e.g., trying to apply it to three numbers. Correction: this identity strictly holds only for exactly two numbers.
- Getting cyclicity length wrong — assuming every digit has a 4-cycle. Correction: 0, 1, 5, 6 have cycle length 1; 4 and 9 have cycle length 2; only 2, 3, 7, 8 have cycle length 4.
- Mishandling remainder 0 in the cyclicity method — when exponent mod cycle-length = 0, the answer is the LAST term of the cycle, not “the 0th term” or zero itself. Example: unit digit of 2^40 → 40 mod 4 = 0 → take the 4th (last) term of {2,4,8,6} = 6, not 0.
- Getting confused by the direction of the alternating sum for divisibility by 11. It does not matter whether you start counting positions from the left or right, as long as you are consistent — but many students mix left-to-right and right-to-left mid-calculation. Pick one direction (right-to-left is the standard convention) and stick to it.
- Wrongly assuming √a + √b = √(a+b). This is FALSE in general (e.g., √4 + √9 = 2+3 = 5, but √13 ≈ 3.6). Surds add like “like terms,” similar to algebraic variables — only when they share the same radical part (e.g., 3√2 + 5√2 = 8√2).
- Forgetting to reduce the exponent correctly for last-two-digit problems — jumping straight to cyclicity of unit digit (which only gives the last ONE digit) when the question asks for the last TWO digits. These require the separate method in Section 9, not the Section 5 method.
- Errors in prime factorization — missing a repeated prime factor (e.g., writing 72 = 2³×3² is correct, but many mistakenly write 72 = 2²×3³ by miscounting). Always verify by multiplying back: 2³×3² = 8×9 = 72 ✓.
- Confusing factors and multiples. Factors of N are numbers that divide N exactly (there are finitely many); multiples of N are numbers N divides into (there are infinitely many).
- Rushing divisibility by 4/8 using digit sum instead of last-two/last-three-digit rule. Divisibility by 4 and 8 has nothing to do with digit sums — that rule is exclusive to 3 and 9.
- Applying the “same remainder → LCM + remainder” shortcut when the problem actually follows the “constant divisor−remainder → LCM − constant” pattern (or vice versa). Always compute divisor−remainder for every given pair first, and only then decide which formula to use (see Trick 12 in Section 13).
- Forgetting that HCF can never exceed the smaller of the two numbers, and LCM can never be smaller than the larger of the two numbers. If your calculated HCF comes out bigger than one of the given numbers, or your LCM comes out smaller than one of them, you have made an arithmetic error — this is a fast built-in sanity check worth using on every HCF/LCM question.
- Stopping trial division too early or too late when testing primality. Remember you only need to test divisibility by primes up to √N (Trick 13) — testing beyond that wastes time, and stopping before that risks missing a valid factor.
- Misreading “exactly divisible” versus “leaves a remainder” in word problems, especially in HCF/LCM-with-remainder questions — always re-read the question once after solving to check whether the remainder condition needs to be added back (as in “smallest number leaving remainder r”) or subtracted first (as in “largest number that divides X and Y leaving remainder r”).
Page 1 of 1