what is the smallest number that when divided by 35, 56 and 91 leaves remainder 7 in each case. Pls answer fast
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Answered by
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3647 is the smallest no.
meghrajsharma:
pls send the calculation as well
ok
Answered by
1
first subtract 7 from each no
35-7=28
56-7=49
91-7=84
now find lcm of 28 ,49 ,84
28=2,2,7
49=7,7
84=2,2,3,7
Lcm = 2x2x7x7x3=588
35-7=28
56-7=49
91-7=84
now find lcm of 28 ,49 ,84
28=2,2,7
49=7,7
84=2,2,3,7
Lcm = 2x2x7x7x3=588
and of 56 is 2*2*2*7
and of 91 is 13*7
and 91= 7x13
5x2x2x2x7x13 =3654
oo sorry
therefore lcm is 7*2*2*2*13*5=3640
3640
and when we add 7 then it will be 3647
remainder remains 7 so, add 7 in 3640=3647
thats it
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