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CAT 2019 Slot 2 — QA Question 26

2 Variable EquationsEasy

In a six-digit number, the sixth, that is, the rightmost, digit is the sum of the first three digits, the fifth digit is the sum of first two digits, the third digit is equal to the first digit, the second digit is twice the first digit and the fourth digit is the sum of fifth and sixth digits. Then, the largest possible value of the fourth digit is

Answer & solution

Correct answer: 7

Answer: 7

Solution

Easy

Let the first digit be aa and write every other digit in terms of aa using the given rules. The fourth digit comes out as a multiple of aa; maximise it subject to staying a single digit (0–9).

1

Express all digits via aa. Let the first digit be aa.

d1=a,d2=2a,d3=a(given)d5=d1+d2=3a(sum of first two)d6=d1+d2+d3=4a(sum of first three)\begin{aligned} &d_1=a,\quad d_2=2a,\quad d_3=a \quad\text{(given)}\\ &d_5 = d_1+d_2 = 3a \quad\text{(sum of first two)}\\ &d_6 = d_1+d_2+d_3 = 4a \quad\text{(sum of first three)} \end{aligned}
2

Fourth digit. It is the sum of the fifth and sixth digits.

d4=d5+d6=3a+4a⇒ d4=7a\begin{aligned} &d_4 = d_5+d_6 = 3a + 4a\\ &\Rightarrow\ d_4 = 7a \end{aligned}
3

Maximise within a single digit. Every digit must be 00–99, so 7a≤97a\le 9 forces a=1a=1.

a=1  ⇒  d4=7⋅1=7\begin{aligned} &a=1 \;\Rightarrow\; d_4 = 7\cdot 1 = 7 \end{aligned}
Largest fourth digit=7\text{Largest fourth digit}=7

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CAT 2019 Slot 2 QA Q26: In a six-digit number, the sixth, that is, the rightmost, digit is the sum of the first three digits, the fift — Solution | TheCATExam