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28 Aug 2008, 16:51
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When the positive integer x is divided by 11, the quotient is y and the remainder 3. When x is divided by 19, the remainder is also 3. What is the remainder when y is divided by 19?
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28 Aug 2008, 19:13
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bigtreezl
A. has to be 0.
since 11 and 19 both are primes and x = 11y+3 = 19k+3, 11y = 19k.
to have 11y = 19k, y must be 19 or its multiple and k has to be 11 and its multiple. for ex:
if y = 19, k = 11
if y = 38, k = 22
and so on. if so, y yeilds 0 reminder when y is divided by 19.
28 Aug 2008, 19:24
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i think 0.
cos , both 11 and 19 are prime nos..
so the no that when divided by both of them , which leaves a remainder 3 is the LCM of 11 and 19 = 209.
so 209+3 = 211 is the reqd no. x
211/11 = 19 + 3=remainder
and 211/19=11+3(remainder)
here, y=19
so 19/19=1 rem=0.
is it OK?
cos , both 11 and 19 are prime nos..
so the no that when divided by both of them , which leaves a remainder 3 is the LCM of 11 and 19 = 209.
so 209+3 = 211 is the reqd no. x
211/11 = 19 + 3=remainder
and 211/19=11+3(remainder)
here, y=19
so 19/19=1 rem=0.
is it OK?
