Question

Find the three consecutive numbers such that if they are divided by 5,7,11 respectively the sum of their quotients will be 9.

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Answer 1

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Let those three consecutive numbers be,

$x, x+1, x+2$

Given that,

$x/5 + (x+1)/7 + (x+2)/11 = 9$

$[77x + 55(x+1) + 35(x+2)]/385 = 9$

$77x + 55x + 55 + 35x + 70 = 3465$

$167x + 125 = 3465$

$167x = 3340$

$x = 3340/167$

$x = 20$

So,

$x = 20$

$x+1 = 21$

$x+2 =22$

Cheers! :)

Answer 2

The answer is given below :

Let us consider that the three consecutive numbers are n, (n+1) and (n+2).

When the three numbers are divided by 5, 7 and 11 respectively, we get three quotients

n/5, (n+1)/7 and (n+2)/11 respectively.

By the given condition,

sum of the quotients = 9

=> n/5 + (n+1)/7 + (n+2)/11 = 9

[Lcm of 5, 7 and 9 is = 5×7×11 = 385]

=> [77n + 55(n+1) + 35(n+2)]/385 = 9

=> (77n + 55n + 55 + 35n + 70) = 9×385

=> 167n + 125 = 3465

=> 167n = 3465 - 125

=> 167n = 3340

=> n = 20

So, the three consecutive numbers are

20, (20+1) and (20+2)

i.e., 20, 21 and 22.

VERIFICATION :

20/5 = 4, quotient = 4

21/7 = 3, quotient = 3

22/11 = 2, quotient = 2

So, the sum of the quotients

= 4 + 3 + 2

= 9.

Thank you, sister for your question.