Question

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Answer 1

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( 1 ) Here HCF of 1001 910 is 91 . Hence among 91 students 1001 pants and 910 can be distributed such that each student get same number of pens and pencils

O R

Total pens = 1001

Total Pencils = 910

We need to find maximum number of students among whom 1001 pens and 910 pencils can be distributed in such a way that each students get SIM number of pants and pencil .

Then we we need to find HCF of 1001 and 9 10

Prime factorization of

1001 = 7×11× 13

910 = 2×5×7

HCF= product of common prime factor of Least Power .

HCF = 7× 13 = 91

Here HCF of 1001 and 910 is 91 .

Hence among 91 students 1001 pens and 910 pencils can be distributed such that each student get same number of pens and pencils .

The Answer is 1001

( 2 ) Required number = H.C.F of ( 1356 - 12 ) , ( 1868 - 12 ) and ( 2764 - 12 )

HCF of 1344 , 1856 , and 2752 = 64

The Answer is 64

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