A heap of shots when made up into groups of 28, 32 and 42 leaves
always a remainder of 5 shots. Find the least number of shots contained
in the heap.
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Answer:
Thus minimum No of shots=672+5
=677
Step-by-step explanation:
The prime factors of 28,32 and 42 are
28= 2x2x7
32=2x2x2x2x2
42=2x3x7
LCM (28,32,42)=2^5 x 3 x 7
=32x21=672
Thus 672 is a smallest number divisible by 28,32 and 42
But in each case 5 shots are left
Thus minimum No of shots=672+5
=677
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