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Find the Smallest Number Which When Increased by 17 is Exactly Divisible by Both 520 and 468. - Mathematics

Numerical

Find the smallest number which when increased by 17 is exactly divisible by both 520 and 468.

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Solution

TO FIND: Smallest number which when increased by 17 is exactly divisible by both 520 and 468.

L.C.M OF 520 and 468

\[520 = 2^3 \times 5 \times 13\]

`468 = 2^2xx3^2xx13`

\[\text{LCM of 520 and 468} = 2^3 \times 3^2 \times 5 \times 13\]

\[ = 4680\]

Hence 4680 is the least number which exactly divides 520 and 468 i.e. we will get a remainder of 0 in this case. But we need the Smallest number which when increased by 17 is exactly divided by 520 and 468.

Therefore

= 4680 -17

= 4663

Hence  4663 is Smallest number which when increased by 17 is exactly divisible by both 520 and 468.

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APPEARS IN

RD Sharma Class 10 Maths
Chapter 1 Real Numbers
Exercise 1.4 | Q 8 | Page 40
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