CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

The supercomputer at Ram Mohan Roy Seminary takes an input of a number N and an X where X is a factor of the number N. In a particular case, N is equal to 83p796161q and x is equal to 11 where 0 < p < q. Find the sum of remainders when N is divided by (p + q) and p successively.

A
6
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
3
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
9
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is D 9
For 83p796161q to be a multiple of 11 (here X is 11), we should have the difference between the sum of odd placed digits and even placed digits as 0 or a multiple of 11.
(8 + p + 9 + 1 + 1) - (3 + 7 + 6 + 6 + q) = (19 + p) - (22 + q). For this difference to be 0, p should be 3 more than q which cannot occur since 0 < p < q. The only way in which (19 + p) - (22 + q) can be a multiple of 11 is if we target a value of - 11 for the expression. One such possibility is if we take p as 1 and q as 9. The number would be 8317961619. On successive division by (p+q) = 10 and 1, the sum of remainders would be 9.

flag
Suggest Corrections
thumbs-up
1
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Factorising Numerator
QUANTITATIVE APTITUDE
Watch in App
Join BYJU'S Learning Program
CrossIcon