Math, asked by prathemesh3437, 4 months ago

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.

Answers

Answered by rkcomp31
1

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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