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ek aisi sankhaya batave jisme 9,11,13 se bhag dene par shesh kramshah 1,2,3 ho?

ek aisi sankhaya batave jisme 9,11,13 se bhag dene par shesh kramshah 1,2,3 ho?

Grade:12th

1 Answers

Nishant Vora IIT Patna
askIITians Faculty 2467 Points
3 years ago
Dear student,

please find the soln below

n/7 = p + 1 as rem

n/9 = q + 2 as rem

n/11 = r + 3 as rem.

Next multiply n by 2 then

2n/7 = p + 2 as rem

2n/9 = q + 4 as rem

2n/11 = r + 6 as rem.

If you add 5 to 2n we get

(2n+5)/7 - (p+1) + 0 rem

(2n+5/9 = (q+1) + 0 as rem

(2n+5)/11 = (r+1) + 0 as rem.

This shows that (2n+5) should be divisible by 7, 9 and 11. Which means that (2n+5) is the LCM of 7, 9 and 11 which is 693.

2n+5 = 693 or

2n = 693–5 = 688, or

n = 344.

The number is thus 344

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