Question

What is the smallest number, which when increased by 7, is divisible by 17,11
and 13?

Answer 1

Answer:

lcm of 17,11,13are 2431

required no.=2431-7=2424

Answer 2

Answer:

Required smallest number is 2424.

Step-by-step explanation:

Given two numbers are 17 and 11 and

Here we want to find smallest number which when increased by 7, is divisible by 17,11 and 13.

Here first need to find LCM of 17,11 and 13.

We know LCM means Lowest Common Multiples.

By some examples we can understand the concept of LCM.

Example -1:

We are taking two numbers 15 and 20.

By prime factorisation,

$15 = 3 \times 5 \\ 20 = 2 \times 2 \times 5$

Therefore lowest common multiples of 15 and 20 is (3×2×2×5) = 60

So,LCM of 15 and 20 is 60.

Example -2:

We are taking two numbers 23 and 30.

By prime factorisation,

$23 = 23 \times 1 \\ 30 = 2 \times 3 \times 5$

Therefore lowest common multiple of 23 and 30 is (23×30) = 690

So,LCM of 23 and 30 is 690.

Here given numbers are 17,11 and 13.

By prime factorisation,

$17 = 17 \times 1 \\ 11 = 11 \times 1 \\ 13 = 13 \times 1$

So lowest common multiple of 17,11 and 13 is $(17 \times 11 \times 13) = 2431$

Therefore LCM of 17,11 and 13 is 2431.

Required number is $(2431 - 7) = 2424$

Therefore,if 2424 increase by 7, is divisible by 17,11 and 13.

Know more about LCM,

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