Question

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38. Certain 3-digit numbers have the
following characteristics
I. All the three digits are different.
II. The number is divisible by 7.
III. The number on reversing the digits
is also divisible by 7.
How many such 3-digit numbers are
there?

(a) 2
(b) 4
(c) 6
(d) 8​

Answer 1

Answer:

option b 4

Step-by-step explanation:

the number be 100x + 10y + z, where x, y and z are single digit numbers and x > 0.

Now the reverse number shall be 100z + 10 y + x. Now it is given that both of them are divisible by 7 and hence their difference shall also be divisible by 7.

Thus (100x + 10y + z) – (100z + 10 y + x) = 7m, where m is an integer.

This implies that 99(x – z) = 7m.

Now since 99 is not divisible by 7 therefore x-z should be divisible by 7. Again since x and z are single digit numbers their difference which is divisible by 7 would actually be 7 only, because for no value of x or z shall x-z be equal to 14 or 21 or 28 etc.

Thus we have x – z = 7 and the values that x and z can take are (9,2) and (8, 1)

Thus the numbers are of the form 1_8 and 2_9.

Now in 1_8, number at unit’s place is 8 so we should get a carry-over of 2 from the division of number formed by the digits at hundred’s and ten’s place to make the number become completely divisible by 7 (we want to make it 28 which is divisible by 7). Thus the ten’s digit should be 6 and number shall be 168. The reverse number 861 is also divisible by 7. Thus two of our required numbers are 168 and 861.

Now in 2_9, number at unit’s place is 9 so we should get a carry-over of 4 from the division of number formed by the digits at hundred’s and ten’s place to make the number become completely divisible by 7 (we want to make it 49 which is divisible by 7). Thus the ten’s digit should be 5 and number shall be 259. The reverse number 952 is also divisible by 7. Thus other two of our numbers are 259 an 952.

Thus there are four numbers satisfying all three conditions and hence the answer is

(b) 4