the least number which when divided by 4, 6, 8, 12 and 16 leaves a remainder of 2 in each case is
50 The least number which when divided by 4, 6, 8, 12, and 16 leaves a remainder of 2 in each case is 50.
Quick Solution Path
This classic LCM problem works like this:
- Find LCM of divisors: Prime factors are 4=2², 6=2×3, 8=2³, 12=2²×3, 16=2⁴. Highest powers give 24×3=482^4\times 3=4824×3=48.
- Subtract remainder: Number - 2 must be divisible by all, so add back 2 to LCM: 48 + 2 = 50.
Verification
Divisor| 50 ÷ Divisor| Quotient| Remainder
---|---|---|---
4| 12.5| 12| 2
6| 8.333...| 8| 2
8| 6.25| 6| 2
12| 4.166...| 4| 2
16| 3.125| 3| 2
No smaller positive integer fits, as next would be 50 + 48k (k>0).
Why It Works
Imagine numbers 2 more than multiples of 48—they're always "2 away" from being evenly divisible by these factors. Popular in exams; multiple sources confirm 50 over options like 46/48/56.
TL;DR: 50 is the answer—LCM(4,6,8,12,16)=48, then +2.
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