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CAT 2021 Slot 2 — QA Question 2

DivisibilityEasy

For a 4-digit number, the sum of its digits in the thousands, hundreds and tens places is 14, the sum of its digits in the hundreds, tens and units places is 15, and the tens place digit is 4 more than the units place digit. Then the highest possible 4-digit number satisfying the above conditions is

Answer & solution

Answer: 4195

Solution

Easy

Name the four digits, turn each statement into an equation, then maximise the number by pushing the leading digits as high as the constraints allow.

1

Translate the conditions. Let the digits (thousands, hundreds, tens, units) be a,b,c,da,b,c,d.

a+b+c=14(thousands+hundreds+tens)…(1)b+c+d=15(hundreds+tens+units)…(2)c=d+4(tens is 4 more than units)…(3)\begin{aligned} &a+b+c=14 \quad\text{(thousands+hundreds+tens)} \quad\dots(1)\\ &b+c+d=15 \quad\text{(hundreds+tens+units)} \quad\dots(2)\\ &c=d+4 \quad\text{(tens is 4 more than units)} \quad\dots(3) \end{aligned}
2

Relate aa and dd. Subtract (1) from (2); the common b+cb+c cancels.

(b+c+d)−(a+b+c)=15−14⇒ d−a=1⇒ a=d−1…(4)\begin{aligned} &(b+c+d)-(a+b+c)=15-14\\ &\Rightarrow\ d-a=1\\ &\Rightarrow\ a=d-1 \quad\dots(4) \end{aligned}
3

Maximise the number. The biggest number needs the biggest leading digit, so we want aa — hence dd — as large as possible. From (3), c=d+4≤9c=d+4\le 9, so d≤5d\le 5. Take d=5d=5.

d=5⇒ c=d+4=9(from (3))⇒ a=d−1=4(from (4))\begin{aligned} &d=5\\ &\Rightarrow\ c=d+4=9 \quad\text{(from (3))}\\ &\Rightarrow\ a=d-1=4 \quad\text{(from (4))} \end{aligned}
4

Find the last digit. Use (1) with a=4, c=9a=4,\ c=9.

b=14−a−c=14−4−9=1\begin{aligned} &b=14-a-c=14-4-9=1 \end{aligned}

So the digits are 4,1,9,54,1,9,5, giving the number 41954195.

Highest possible number=4195\text{Highest possible number}=4195
CAT 2021 Slot 2 QA Q2: For a 4-digit number, the sum of its digits in the thousands, hundreds and tens places is 14, the sum of its d — Solution | TheCATExam