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CAT 2021 Slot 1 — QA Question 19

2 Variable EquationsEasy

How many three-digit numbers are greater than 100 and increase by 198 when the three digits are arranged in the reverse order?

Answer & solution

Answer: 70

Solution

Easy

Write the three-digit number in place value as 100a+10b+c100a+10b+c and its reverse as 100c+10b+a100c+10b+a. Translate "increases by 198198 on reversal" into an equation; the middle digit cancels, leaving a simple condition on aa and cc. Then count the valid digit choices.

1

Set up the equation. Let the number be abc‾=100a+10b+c\overline{abc}=100a+10b+c (with a≥1a\ge1). Its reverse is cba‾=100c+10b+a\overline{cba}=100c+10b+a.

(100a+10b+c)+198=100c+10b+a\begin{aligned} &(100a+10b+c)+198=100c+10b+a \end{aligned}
2

Simplify. The 10b10b terms cancel.

198=99c−99a⇒ c−a=2\begin{aligned} &198=99c-99a\\ &\Rightarrow\ c-a=2 \end{aligned}
3

Count valid (a,c)(a,c) pairs. We need a≥1a\ge1 (leading digit) and c≤9c\le9 with c=a+2c=a+2.

a=1,…,7 ⇒ (a,c)=(1,3),(2,4),…,(7,9)⇒ 7 pairs\begin{aligned} &a=1,\dots,7\ \Rightarrow\ (a,c)=(1,3),(2,4),\dots,(7,9)\\ &\Rightarrow\ 7\ \text{pairs} \end{aligned}
4

Account for the free middle digit. bb is unconstrained, taking any of 0,1,…,90,1,\dots,9 — that is 1010 choices for each pair.

7×10=70\begin{aligned} &7\times10=70 \end{aligned}
70 numbers70\ \text{numbers}
Need a hint?

Reversing changes only the hundreds and units digits, so the change is 99(c−a)99(c-a). Set 99(c−a)=19899(c-a)=198 to get c−a=2c-a=2, then count digit choices.

CAT 2021 Slot 1 QA Q19: How many three-digit numbers are greater than 100 and increase by 198 when the three digits are arranged in th — Solution | TheCATExam