[3]
27
Section c
What is the smallest number that, when divided by 35, 56 and 91 leaves remainders of 7 in
each case?
OR
Find the HCF oſ 65 and 117 and find a pair of integral values of m and n such that HCF = 65m -
117n.
that one and only one out ofn, (n + 2) or (n + 4) is divisible by 3, where n EN.
13
Answers
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1
Answer:
Answer of 1
7 is the smallest number required
Step-by-step explanation:
35-7=28
56-7=49
91-7=84
28=2*2*7
49=7*7
84=2*2*3*7
hcf = 7
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