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CAT 2017 Slot 2 — QA Question 30

Numbers (P&C)Easy

How many four digit numbers, which are divisible by 6, can be formed using the digits 0, 2, 3, 4, 6, such that no digit is used more than once and 0 does not occur in the left-most position?

Answer & solution

Correct answer: 50

Answer: 50

Solution

Easy

Divisible by 66 means divisible by both 22 (last digit even) and 33 (digit sum a multiple of 33). First pick which four of the five digits are used so the sum is a multiple of 33, then count valid arrangements in each case, barring 00 from the first place.

1

Choose digit sets with sum divisible by 33. The digits available are 0,2,3,4,60,2,3,4,6 (total 1515). Dropping one digit, the chosen four must sum to a multiple of 33. The valid four-digit sets are:

{0,2,3,4} (sum 9),{0,2,4,6} (sum 12),{2,3,4,6} (sum 15)\begin{aligned} &\{0,2,3,4\}\ (\text{sum }9),\quad \{0,2,4,6\}\ (\text{sum }12),\quad \{2,3,4,6\}\ (\text{sum }15) \end{aligned}
2

Case {0,2,3,4}\{0,2,3,4\}. The last digit must be even (0,2,40,2,4) and the first digit cannot be 00. Counting by first digit:

first=2: last∈{0,4}, middle two arrange ⇒ 4first=4: last∈{0,2}, middle two arrange ⇒ 4first=3: last∈{0,2,4}, remaining arrange=3! ⇒ 6⇒ 4+4+6=14\begin{aligned} &\text{first}=2:\ \text{last}\in\{0,4\},\ \text{middle two arrange}\ \Rightarrow\ 4\\ &\text{first}=4:\ \text{last}\in\{0,2\},\ \text{middle two arrange}\ \Rightarrow\ 4\\ &\text{first}=3:\ \text{last}\in\{0,2,4\},\ \text{remaining arrange}=3!\ \Rightarrow\ 6\\ &\Rightarrow\ 4+4+6=14 \end{aligned}
3

Case {0,2,4,6}\{0,2,4,6\}. Every digit is even, so the last digit is automatically even; the only constraint is that the first digit is not 00.

total arrangements=4!=24arrangements with leading 0=3!=6⇒ 24−6=18\begin{aligned} &\text{total arrangements}=4!=24\\ &\text{arrangements with leading }0=3!=6\\ &\Rightarrow\ 24-6=18 \end{aligned}
4

Case {2,3,4,6}\{2,3,4,6\}. No 00, so the only constraint is an even last digit; it must be one of 2,4,62,4,6 (33 choices), and the other three digits fill the remaining places.

last∈{2,4,6} ⇒ 3 choicesremaining 3 digits arrange=3!=6⇒ 3×6=18\begin{aligned} &\text{last}\in\{2,4,6\}\ \Rightarrow\ 3\ \text{choices}\\ &\text{remaining 3 digits arrange}=3!=6\\ &\Rightarrow\ 3\times 6=18 \end{aligned}
5

Add the cases.

14+18+18=50\begin{aligned} &14+18+18=50 \end{aligned}
Number of four-digit numbers=50\text{Number of four-digit numbers}=50

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CAT 2017 Slot 2 QA Q30: How many four digit numbers, which are divisible by 6, can be formed using the digits 0, 2, 3, 4, 6, such that — Solution | TheCATExam