CAT 2018 Slot 2 — QA Question 25

Number TheoryEasy

How many two-digit numbers, with a non-zero digit in the units place, are there which are more than thrice the number formed by interchanging the positions of its digits?

Answer & solution

Correct answer: 6

  • 6

  • B

    8

  • C

    7

  • D

    5

Solution

Easy

Write the number as 10x+y10x+y and its reverse as 10y+x10y+x. Translate “more than thrice the reversed number” into an inequality, then count the digit pairs (x,y)(x,y) with y≠0y\ne 0 that satisfy it.

1

Set up the inequality. Let the number be 10x+y10x+y (tens digit xx, units digit yy, x≥1x\ge 1, y≥1y\ge 1).

10x+y>3(10y+x)⇒ 10x+y>30y+3x⇒ 7x>29y(collect terms)⇒ x>297 y\begin{aligned} &10x+y > 3(10y+x)\\ &\Rightarrow\ 10x+y > 30y+3x\\ &\Rightarrow\ 7x > 29y \quad\text{(collect terms)}\\ &\Rightarrow\ x > \tfrac{29}{7}\,y \end{aligned}
2

Count for each units digit yy. Here 297≈4.14\tfrac{29}{7}\approx 4.14, and xx must be a single digit ≤9\le 9.

y=1: x>4.14⇒x∈{5,6,7,8,9}(5 values)y=2: x>8.29⇒x∈{9}(1 value)y≥3: x>12.4⇒no single digit(0 values)\begin{aligned} &y=1:\ x>4.14 \Rightarrow x\in\{5,6,7,8,9\} \quad\text{(5 values)}\\ &y=2:\ x>8.29 \Rightarrow x\in\{9\} \quad\text{(1 value)}\\ &y\ge 3:\ x>12.4 \Rightarrow \text{no single digit} \quad\text{(0 values)} \end{aligned}
3

Total. Add the cases.

5+1=6\begin{aligned} &5 + 1 = 6 \end{aligned}
6\textbf{6}

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CAT 2018 Slot 2 QA Q25 — Number Theory solution | TheCATExam