Use place values to find a number · source reference

Printed question

$N$ is a three-digit positive number that is 26 times the sum of its digits. When the digits of $N$ are reversed, the new number is 198 more than $N$. Given that the digits of $N$ are 3 consecutive terms of an arithmetic progression, find $N$. [4]

[2014/MJC/mid year/1]

Printed working

Let $N$ be the 3-digit number $\overline{abc}$, where $a,b,c\in\mathbb Z$, $1\le a,c\le9,0\le b\le9$.

$\overline{abc}=26(a+b+c)$.

$100a+10b+c=26(a+b+c)$.

$74a-16b-25c=0$ — (1).

$\overline{cba}-\overline{abc}=198$.

$100c+10b+a-100a-10b-c=198$.

$-a+c=2$ — (2).

The 3 digits are consecutive terms of a AP:

$b-a=c-b$.

$-a+2b-c=0$ — (3).

By GC, $a=2,b=3,c=4$. Thus $N$ is the number 234.

Printed alternative solution

Let $a$ be the first digit of $N$ and $d$ be the common difference between the digits.

$100a+10(a+d)+(a+2d)=N$.

$26[a+(a+d)+(a+2d)]=N$.

$a+10(a+d)+100(a+2d)-N=198$.

Simplifying,

$111a+12d-N=0$.

$78a+78d-N=0$.

$111a+210d-N=198$.

Using GC, $a=2,d=1,N=234$.

$\therefore N=234$.

Teacher notes

Example: $235=100(2)+10(3)+5$. So $N=\overline{abc}=100(a)+10(b)+c$.

AP: $a,b,c$. Common difference $=b-a=c-b$.

Printed answer key

Source provenance

2025 TJC H2 Math (9758) JC2 MSM: EQUATIONS. Question sheet printed pages 15–19; solution sheet printed pages 1–12. Original exam attribution is retained above. Currency dollar signs are spelled out outside maths to avoid confusing currency with TeX delimiters.