Unique Rule

Number Theory Level pending

Assume that $10a+b$ is a 2-digit integer, where $a$ and $b$ are positive integers. If the positive difference between the integer and the sum of its digits can be expressed in the form of $10m+n$ , solve for the positive value of $S$ in which $S^2 = m+n$ .

Additional Note: $0 < a,b,m,n < 10$

Challenge : Prove that for any 2-digit integer, having the same rule as above, the value of $S^2$ is always the same.

The answer is 3.

1 solution

Nathan Laurence
Dec 4, 2016

Solution for the first question

Let us assume that the number is $28$ , therefore $a = 2, b = 8$ .

28 - (2 + 8) = 18 [We get $m = 1, n = 8$ ]

$S^2 = m+n$

$S^2 = 9$

$S = 3$ , $S = -3$

But because the question says the positive value of $S$ , $S = 3$

Challenge Solution

From the information, we get the equation:

$(10a+b) - (a+b) = (10m+n)$

[Simplify] $9a = 10m+n$

$9a = m+n+9m$

$9a = S^2+9m$

$S^2 = 9(a-m)$

Now we need to solve for the value of $a-m$ ,

The maximum possible number is 99, where $90+9-9-9 = 81$

The minimum possible number is 11, where \(10+1-1-1 = 9)

If we observe, the difference between the 'ones' digit of the first (a) and the second (m) has a range of $1-9$, and does not exceeds 9 (e.g. 10). Thus, we can conclude that the difference between the 'tens' digit is always $1$

Finally, we get $S^2 = 9(1)$ -> $S^2 = 9$ [PROVEN]

What makes this question interesting, is that the property holds for all two-digit numbers. I encourage you to post another problem to reflect that.

Calvin Lin Staff - 4 years, 6 months ago

Unique Rule

The answer is 3.

1 solution

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