10. Solved Miscellaneous Examples (Mixed Difficulty)

These examples mix multiple ideas from this chapter the way an actual exam paper does — read each one and identify which rule/formula is being triggered before checking the solution.

Example 1: What least number must be subtracted from 1,852 so that the remainder is exactly divisible by 7? 1852 ÷ 7: 7×264=1848, remainder =1852-1848=4. Least number to subtract = 4 (giving 1848, which is 7×264).

Example 2: In the 7-digit number 5 x 4 8 2 3 y, if the number is divisible by 88, find the value of x+y. 88=8×11. Testing systematically (or by the standard combined 8-and-11 method): the last three digits “23y” must be divisible by 8, and the alternating digit sum must be divisible by 11. Working through both conditions together gives the unique solution x=2, y=2, making the number 5,248,232. Check: 5248232 ÷ 88 = 59639 exact. ✓ x+y = 2+2 = 4.

Example 3: Find the total number of factors of 5 contained in 1000! (i.e., how many times can 1000! be divided by 5). Apply Legendre’s Formula with p=5: ⌊1000/5⌋=200, ⌊1000/25⌋=40, ⌊1000/125⌋=8, ⌊1000/625⌋=1, ⌊1000/3125⌋=0. Total =200+40+8+1=249. (This is also, by definition, the number of trailing zeros in 1000!.)

Example 4: If a number N is divisible by both 15 and 20, which of the following must also divide N: 30, 45, 60, or 90? LCM(15,20)=60 (since 15=3×5, 20=4×5, LCM takes highest powers: 4×3×5=60). Since N is a multiple of 60, it is automatically a multiple of every factor of 60 — and among the options, 60 itself is guaranteed; 30 is also a factor of 60 so it’s guaranteed too, but 45 and 90 are not necessarily factors of every multiple of 60 (e.g., N=60 itself is not divisible by 45 or 90). Answer: N must be divisible by 60 (and by 30, since 30 | 60).

Example 5: A trader has 1,890 identical items which he wants to pack into boxes such that each box contains the same number of items, with more than 1 and fewer than 100 items per box. Using divisibility, how many valid box sizes are possible? This requires finding all factors of 1890 between 2 and 99. 1890 = 2×3^3×5×7. Its factors in that range: 2,3,5,6,7,9,10,14,15,18,21,27,30,35,42,45,54,63,70,90. Counting these: 20 valid box sizes.

Example 6: Find the remainder when 50! is divided by 53 (53 is prime). This uses Wilson’s Theorem: for a prime p, (p-1)! ≡ -1 (mod p). Here we need 50!, not 52!, so first relate them: 52! = 52×51×50!, and 52!≡-1(mod 53) by Wilson’s theorem. Since 52≡-1 and 51≡-2(mod 53), we get (-1)(-2)×50! ≡ -1 (mod 53), i.e., 2×50!≡-1(mod 53), i.e., 2×50!≡52(mod 53), so 50!≡26(mod 53). Remainder = 26. (This Wilson’s-theorem style question is more of a CGL Tier-II / advanced-level curiosity — most SSC/RRB papers stay within the simpler Legendre’s Formula and rule-based questions covered earlier in this chapter.)

Example 7 (factor-counting within a range, harder version of Example 5): How many factors of 2,520 lie strictly between 10 and 100? 2520 = 2^3×32×5×7. Listing all its factors and keeping only those strictly between 10 and 100: 12, 14, 15, 18, 20, 21, 24, 28, 30, 35, 36, 40, 42, 45, 56, 60, 63, 70, 72, 84, 90. Counting these: 21 factors.

Example 8 (divisibility of a large expression via a remainder identity): Find the remainder when 17^(23) is divided by 18. Notice 17 ≡ -1 (mod 18). So 17^(23) ≡ (-1)^(23) = -1 (mod 18). Since -1 ≡ 17 (mod 18), the remainder is 17. This is a direct application of the same a^n± b^n reasoning from Section 3, disguised as a remainder question: writing 17=18-1, we have 17⁽²³⁾⁺¹(23) = 17^(23)+1, and since the exponent 23 is odd, (18-1)^(23)+1 is divisible by 18-1+1=18… more directly, 17^(23)-(-1)(23) ·(-1)=17^(23)+1 is divisible by 17-(-1)=18 (Rule 3 of Section 3, with a=17,b=1, odd n), which is exactly the statement that 17^(23)≡-1≡17(mod 18).