JIPMAT 2024 (QA) - What will be the least number which when divided by 12, 21 and 35 leaves 6 as a remainder in each case? | PYQs + Solutions | AfterBoards
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JIPMAT 2024

Number System
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Remainder

Easy

What will be the least number which when divided by 12, 21 and 35 leaves 6 as a remainder in each case?

Correct Option: 1
Let's say the required number is nn. \newline When nn is divided by 1212, 2121 and 3535, remainder is 66 in each case. \newline
This means: \newline n=12k1+6n = 12k_1 + 6 \newline n=21k2+6n = 21k_2 + 6 \newline n=35k3+6n = 35k_3 + 6 \newline where k1k_1, k2k_2, k3k_3 are some integers \newline
Therefore, n−6n - 6 is divisible by 1212, 2121 and 3535. \newline In other words, n−6n - 6 is divisible by LCM(12,21,35)LCM(12, 21, 35) \newline
Let's find LCM(12,21,35)LCM(12, 21, 35): \newline 12=22×312 = 2^2 × 3 \newline 21=3×721 = 3 × 7 \newline 35=5×735 = 5 × 7 \newline Therefore, LCM(12,21,35)=420LCM(12, 21, 35) = 420 \newline
So, n−6=420kn - 6 = 420k, where kk is a non-negative integer \newline The least value of nn will be when k=1k = 1
Therefore, n=420+6=426n = 420 + 6 = 426.

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