Question

In each of the following numbers, replace * by the smallest number to make it divisible by 11:
(i) 26*5
(ii) 39*43
(iii) 86*72
(iv) 467*91
(v) 1723*4
(vi) 9*8071

Answer 1

(i) 26*5
Sum of the digits at odd places = 5 + 6 = 11
Sum of the digits at even places = * + 2
Difference = sum of odd terms – sum of even terms
= 11 – (* + 2)
= 11 – * – 2
= 9 – *
Now, (9 – *) will be divisible by 11 if * = 9.
i.e., 9 – 9 = 0
0 is divisible by 11.
∴ * = 9
Hence, the number is 2695.

(ii) 39*43
Sum of the digits at odd places = 3 + * + 3 = 6 + *
Sum of the digits at even places = 4 + 9 = 13
Difference = sum of odd terms – sum of even terms
= 6 + * – 13
= * – 7
Now, (* – 7) will be divisible by 11 if * = 7.
i.e., 7 – 7 = 0
0 is divisible by 11.
∴ * = 7
Hence, the number is 39743.

(iii) 86*72
Sum of the digits at odd places 2 + * + 8 = 10 + *
Sum of the digits at even places 6 + 7 = 13
Difference = sum of odd terms – sum of even terms
= 10 + * – 13
= * – 3
Now, (* – 3) will be divisible by 11 if * = 3.
i.e., 3 – 3 = 0
0 is divisible by 11.
∴ * = 3
Hence, the number is 86372.

(iv) 467*91
Sum of the digits at odd places 1 + * + 6 = 7 + *
Sum of the digits at even places 9 + 7 + 4 = 20
Difference = sum of odd terms – sum of even terms
= (7 + *) − 20
= * − 13
Now, (* −13) will be divisible by 11 if * = 2.
i.e., 2− 13 = −11
−11 is divisible by 11.
∴ * = 2
Hence, the number is 467291.

(v) 1723*4
Sum of the digits at odd places 4+ 3+ 7= 14
Sum of the digits at even places *+2+1 = 3 + *
Difference = sum of odd terms – sum of even terms
= 14 – (3 + *)
= 11 − *
Now, (11 − *) will be divisible by 11 if * = 0.
i.e., 11 − 0 = 11
11 is divisible by 11.
∴ * = 0
Hence, the number is 172304.

(vi) 9*8071
Sum of the digits at odd places 1+0+* = 1 + *
Sum of the digits at even places 7 + 8 + 9 = 24
Difference = sum of odd terms – sum of even terms
=1 + * – 24
= * − 23
Now, (* − 23) will be divisible by 11 if * = 1.
i.e., 1 − 23 = −22
−22 is divisible by 11.
∴ * = 1
Hence, the number is 918071.