How many pairs of positive integers m, n satisfy: 1m+4n=112, where n is an odd integer less than 60?
A
5
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
3
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
6
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
4
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is B 3 1m=112−4n⇒m=12nn−48 Since m is positive, n must be greater than 48. Possible odd values of n such that 48 < n < 60 are 49, 51, 53, 55, 57 and 59, and 49, 51, and 57 give integral values of m. Hence option (b) is the answer.