Find the largest two-digit number which leaves the same remainder on dividing 2614 and 2458.
Answers
Answer:
A number N when divided by a number n , it can be factorized as following:
N = nq + rN=nq+r ;
Where N is the number, n is the divisor , q is quotient and r is the remainder.
1)2614 = n()q_{1} +r2614=n()q
1
+r
2)2458= n(q_{2}) + r2458=n(q
2
)+r
substract equation 1 from 2 ;
2614-2458 = n(q_{1}-q_{2} )2614−2458=n(q
1
−q
2
) ;
156= n(q_{1} -q_{2})156=n(q
1
−q
2
) ;
\begin{gathered}156 = (2)(2)(3)(13)\\\end{gathered}
156=(2)(2)(3)(13)
We need to find the largest two digit number which satisfies the above equation
possible values of n = 13,2,3,6,39,78,52,156
out of these highest two digit number is 72
The largest two digit number which leaves same remainder on dividing 2614 and 2458 is 72
Answer:
72
Explanation:
for the first one line