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CAT 2024 Slot 1 — QA Question 5

Number TheoryEasy

The sum of all four-digit numbers that can be formed with the distinct non-zero digits a, b, c and d, with each digit appearing exactly once in every number, is 153310 + n, where n is a single digit natural number. Then, the value of (a + b + c + d + n) is

Answer & solution

Answer: 31

Solution

Hard

With four distinct digits permuted, each digit lands in each place value an equal number of times. That gives a clean formula for the grand total in terms of (a+b+c+d)(a+b+c+d). Match it to 153310+n153310+n to pin down both the digit sum and nn.

1

Count appearances per place. There are 4!=244!=24 arrangements. Fix one digit in one place: the other three fill the remaining places in 3!=63!=6 ways. So each digit appears 66 times in each of the thousands, hundreds, tens and units places.

appearances of each digit per place=3!=6\begin{aligned} &\text{appearances of each digit per place}=3!=6 \end{aligned}
2

Build the total sum. Each place contributes 6(a+b+c+d)6(a+b+c+d) times its place value; place values add to 11111111.

Sum=6(a+b+c+d) (1000+100+10+1)⇒ Sum=6⋅1111⋅(a+b+c+d)(from step 1)⇒ Sum=6666 (a+b+c+d)\begin{aligned} &\text{Sum}=6(a+b+c+d)\,(1000+100+10+1)\\ &\Rightarrow\ \text{Sum}=6\cdot 1111\cdot(a+b+c+d) \quad\text{(from step 1)}\\ &\Rightarrow\ \text{Sum}=6666\,(a+b+c+d) \end{aligned}
3

Match to 153310+n153310+n. Since nn is a single-digit natural number (11–99), find the multiple of 66666666 just above 153310153310.

6666×23=153318⇒ 153318=153310+8(so n=8)⇒ a+b+c+d=23(matching multiplier)\begin{aligned} &6666\times 23=153318\\ &\Rightarrow\ 153318=153310+8 \quad\text{(so }n=8\text{)}\\ &\Rightarrow\ a+b+c+d=23 \quad\text{(matching multiplier)} \end{aligned}
4

Combine. Add the digit sum from step 3 to nn.

a+b+c+d+n=23+8=31\begin{aligned} &a+b+c+d+n=23+8=31 \end{aligned}
a+b+c+d+n=31a+b+c+d+n=31

Any sum of all permutations of kk distinct digits equals (k−1)!×(repunit of k ones)×(digit sum)(k-1)!\times(\text{repunit of }k\text{ ones})\times(\text{digit sum}). Here 3!×1111×(sum)=6666×(sum)3!\times1111\times(\text{sum})=6666\times(\text{sum}), so the total is always a multiple of 66666666 — just divide.

CAT 2024 Slot 1 QA Q5: The sum of all four-digit numbers that can be formed with the distinct non-zero digits a, b, c and d, with eac — Solution | TheCATExam