Question Number 104413 by Anindita last updated on 21/Jul/20

The sum of two integers is 304  and their GCD is 19. What are the  numbers?

Answered by Rasheed.Sindhi last updated on 22/Jul/20

Integers: 19x,19y  Sum: 19x+19y=304                     x+y=16                           y=16−x    Required integers are 19x,19(16−x)  ∀ x∈Z, with GCD(x,16−x)=1

Commented byfloor(10²Eta[1]) last updated on 21/Jul/20

you are including all the cases when  gcd(x, y)≠1 which is not correct

Commented byRasheed.Sindhi last updated on 22/Jul/20

Yes this restriction I forgot.  Thank you!

Answered by 1549442205PVT last updated on 21/Jul/20

Since GCD(a,b)=19 ⇒ { ((a=19m)),((b=19n)) :}  where gcd(m,n)=1  From the hypothesis a+b=304 we get  19(m+n)=304⇒m+n=16.  WLOG we suppose m<n.Then we have  the following cases:  (note that gcd(m,n)=1⇒m is odd)  i)m=1⇒n=15⇒(a,b)=(19,285)  ii)m=3⇒n=13⇒(a,b)=(57,247)  iii)m=5⇒n=11⇒(a,b)=(209,95)  iv)m=7⇒n=9⇒(a,b)=(153,151)  Thus all the pairs (a,b) of positive  integers which we need to find as:  (a,b)∈{(19;285),(57,247),(209;95),(153;151)}  and the permutations of them

Commented byfloor(10²Eta[1]) last updated on 21/Jul/20

gcd(153,151)=1≠19.  the last case should be  (a,b)=(133,171)

Commented byfloor(10²Eta[1]) last updated on 21/Jul/20

m=2s+1, n=2t+1  m+n=2s+2t+2=16⇒s+t=7  s=k, t=7−k  m=2k+1, n=15−2k  2k+1<15−2k⇒k≤3  so all the pairs (a, b) of integers are:  (a, b)={19(2k+1), 19(15−2k)} ∀ k≤3

Commented by1549442205PVT last updated on 22/Jul/20

Thank you sir,i mistaked this case