Question

Given 63.63 = m(21 = n/100), m, n positive integers with n < 100. The value of (m+n) is:
A.21
B. 24
(C) 104
(D) 101​

Answer 1

Answer:

This implies that

x2+2ax=4x−4a−13

or

x2+2ax−4x+4a+13=0

or

x2+(2a−4)x+(4a+13)=0

Since the equation has just one solution instead of the usual two distinct solutions, then the two solutions must be same i.e. discriminant = 0.

Hence we get that

(2a−4)2=4⋅1⋅(4a+13)

or

4a2−16a+16=16a+52

or

4a2−32a−36=0

or

a2−8a−9=0

or

(a−9)(a+1)=0

So the values of a are −1 and 9.

Answer 2

Answer:

24

Step-by-step explanation:

6363 = m ( 2100 + n ) / 100

m = 6363 / 2100 + n

6363 = 3 x 2121 / 2100 + n

So 6363 should be divisible by 2100 + n

2121 = 2100 + n

n = 21

m = 3

m + n = 21 + 3 = 24

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