Question 1 Real Numbers Exercise 1.4
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Answer:
First, let’s find the smallest number which is exactly divisible by all 35, 56, and 91.
Which is simply just the LCM of the three numbers. By prime factorization, we get
35 = 5 × 7
56 = 23 × 7
91 = 13 × 7
∴ L.C.M (35, 56 and 91) = 23 × 7 × 5 × 13 = 3640
Hence, 3640 is the smallest number which is exactly divisible 28, 42 and 84 i.e. we will get a remainder of 0 in each case. But, we need the smallest number that when divided by 35, 56, and 91 leaves the remainder of 7 in each case.
So that is found by,
3640 + 7 = 3647
∴ 3647 should be the smallest number that when divided by 35, 56, and 91 leaves the remainder of 7 in each case.
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TO FIND: Smallest number that, when divided by 35, 56 and 91 leaves remainder of 7 in each case
L.C.M OF 35, 56 and 91
`35= 5xx7`
`56=2^2xx7`
`91=13xx7`
L.C.M of 35,56 and 91 = `2^2xx5xx7xx13`
=3640
Hence 84 is the least number which exactly divides 28, 42 and 84 i.e. we will get a remainder of 0 in this case. But we need the smallest number that, when divided by 35, 56 and 91 leaves remainder of 7 in each case
Therefore
= 3640 +7
= 3640
Hence 3640 is smallest number that, when divided by 35, 56 and 91 leaves remainder of 7 in each case.