Question

A number of two digits has 3 for its units digit and the sum of digits is 1/7 of the number itself. What is the number ?​

Answer 1

Answer:

A number of two digits have 3 for its units digit, and the sum of digits is 17 of the number itself. The number is a. 43 b. 63 c.

Step-by-step explanation:

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Answer 2

Answer:

Clearly, the answer for a 2 digit number is 63 as it is the only 2 digit number divisible by 7 that has 3 in its decimal form. Hence, for 2 digit numbers we don’t need to know that the sum of the digits is one seventh of the number itself, only that the number is divisible by 7. The fact that 63 = 7x(6+3) is almost an afterthought that we still have to verify to complete the question.

A more interesting question is whether there are numbers with more digits for which this holds.

If our number is, in decimal notation a1a2…an3 , then 7 times the sum of the digits is at most 7(9n+3)=63n+21 which has to be greater or equal to 10n . This is possible for n=1 and n=2 and impossible for all other values of n . So we are looking for either a 2 digit or 3 digit number. For 2 digit numbers, we already checked that the only possibility is 63.

We are now trying to solve 7(x+y+3)=100x+10y+3 where the number we are looking for would be, in decimal form, xy3 . This is a Diophantine equation that reduces to 93x=3y−18 or 31x=y−6 . Since x and y are digits, the only possibility is x=0,y=6 which indirectly also proves that the only 2 digit number is 63.