Find the two-digit number whose digits add to 7 and whose number of tens minus its number of units is 43 times smaller than the number itself.

Step-by-step solution

1. 1

   Give the digits names and write the number in place-value form. Let $a$ be the tens digit and $b$ the units digit, with $1\le a\le 9$ and $0\le b\le 9$. The number itself is

   $N=10a+b,$

   not $ab$ — this expansion is what turns a digit puzzle into ordinary algebra.

2. 2

   Translate both sentences into equations. The digit sum gives $a+b=7$. 'The difference $a-b$ is $43$ times smaller than $N$' means $N$ is $43$ times as large as that difference:

   $10a+b=43(a-b).$

3. 3

   Simplify the second equation.

   $10a+b=43a-43b\ \Longrightarrow\ 44b=33a\ \Longrightarrow\ 4b=3a.$

   Dividing by the common factor $11$ is what makes the relationship readable: the digits are in the ratio $a:b=4:3$.

4. 4

   Combine with the digit-sum condition. From $4b=3a$ we get $a=\tfrac{4b}{3}$, and substituting into $a+b=7$:

   $\frac{4b}{3}+b=7\ \Longrightarrow\ \frac{7b}{3}=7\ \Longrightarrow\ b=3,\qquad a=4.$

5. 5

   Assemble and verify the number. The number is $N=10(4)+3=43$. Check both conditions: the digit sum is $4+3=7$; the digit difference is $4-3=1$, and $43\div 1=43$, so the difference really is $43$ times smaller. A brute-force scan of all two-digit numbers confirms $43$ is the only solution.