Week 8

Example 1. Find all two-digit numbers such that each is divisible by the product of its two digits.

Example 2. In January 1993, Shyam’s age equaled the sum of digits comprised in her birth year. What year was Shyam born in?

Of course, you can find the solution just below, but it is highly recommended that, you first try to solve it on your own.

Just remember the words of Paul Halmos, who says “the only way to learn mathematics is to do mathematics”.

Solution 1. Let 10p + q (= pq) be the required two-digit number. Note the condition that we can impose on p and q. As per the question, there exist a positive integer t such that 10p + q = tpq.

Then q = p(tq − 10), so that p | q. Hence 0 < pq and p ≤ 4 (otherwise q ≥ 10).

If p = q, then 10p + q = tp2, so that tp = 11 = 11 × 1. Since 11 is a prime, we take p = 1 = q to get 11 as a solution.

When p = 4, then q = 8. However, no positive integer will satisfy 48 = 32t, so p ≠ 4.

When p = 3, then 30 = q(3t − 1). Since q ≤ 9, q | 30 and 3 | q, so q = 6. The number in this case is 36 = 2 × 3 × 6, which is another solution.

When p = 2, then 20 = q(2t − 1). Since q ≤ 9, q | 20 and 2 | q, so q = 4. The number becomes 24 = 3 × 2 × 4, which is another solution.

When p = 1, then 10 = q(t − 1).  Since q ≤ 9, q | 10 and 1 | q, so q = 2 or 5. We find two numbers 12 and 15 in this case, which are solutions also.

Therefore, the solutions are 11, 12, 15, 24 and 36.

Solution 2. The sum of digits of the year earlier than 1993 equals no more than 1+ 9 + 9 + 9 = 28, so that Shyam is at most 28 years old. This means that Shyam was born in 1993 − 28 = 1965, at the earliest. The sum of digits of any year between 1965 and 1993 equals at least 1 + 9 + 6 + 0 = 16, thus Shyam was born at the latest in 1993 − 16 = 1977. Let us consider two cases:

Case 1. Shyam was born in the sixties, i.e., in the year 1960 + x, where x is a single-digit number.

The sum of the digits of the year in which he was born equals 1 + 9 + 6 + x = 16 + x.

Shyam would be 16 + x years old in the year 1960 + x + 16 + x = 1976 + 2x, i.e., this year is expressed by an even number.

However, it transpires from the question that Shyam would be 16 + x years in the year 1993, which is an odd number. In that case, Shyam can’t have been born in the sixties.

Case 2. Shyam was born in the seventies, i.e., in the year 1970 + x, where x is a single digit number.

The sum of the digits of the year he was born in is 1 + 9 + 7 + x = 17+ x.

Thus Shyam would be 17 + x years old in the year 1970 + x + 17 + x = 1987 + 2x. But as per the question, 1987 + 2x = 1993, whence 2x = 1993 − 1987 = 6.

Thus, x = 3, so the year to find is 1973.

Therefore, Shyam was born in 1973.

3 comments

  1. Here is a shorter solution:
    1993=1900+10a+b+(1+9+a+b)=1900+10+a+b(year of birth:19ab)
    Then:1993=1910+11a+2b
    1993-1910=11a+2b
    83=11a+2b
    83-2b=11a
    11|(83-2b) b€{0,1,2,3,4,5,6,7,8,9}
    Thé only value of b is 3
    Then 11a=77and :a=7
    Thé year of birth is:1973

    Liked by 1 person

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