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CAT 2018 Slot 1 — QA Question 27

DivisibilityEasy

The number of integers x such that 0.25 ≤ 2x ≤ 200, and 2x + 2 is perfectly divisible by either 3 or 4, is 

Answer & solution

Correct answer: 5

Answer: 5

Solution

Easy

First find which integers xx keep 2x2^x in the range [0.25, 200][0.25,\,200]. Then, for each such xx, test whether 2x+22^x+2 is divisible by 33 or 44 and count the successes.

1

Find the allowed integer values of xx. Write the bounds as powers of 22.

0.25≤2x≤200⇒ 2−2≤2x≤200(0.25=2−2)⇒ −2≤x≤7(27=128≤200<256=28)\begin{aligned} &0.25 \le 2^x \le 200\\ &\Rightarrow\ 2^{-2} \le 2^x \le 200 \quad\text{(}0.25=2^{-2}\text{)}\\ &\Rightarrow\ -2 \le x \le 7 \quad\text{(}2^7=128\le 200<256=2^8\text{)} \end{aligned}

So x∈{−2,−1,0,1,2,3,4,5,6,7}x\in\{-2,-1,0,1,2,3,4,5,6,7\}.

2

Test divisibility of 2x+22^x+2 by 33 or 44. For x=−2,−1x=-2,-1, 2x+22^x+2 is not an integer, so discard them. Check the rest.

x=0: 20+2=3(div by 3) ✓x=1: 21+2=4(div by 4) ✓x=2: 22+2=6(div by 3) ✓x=3: 23+2=10(neither)x=4: 24+2=18(div by 3) ✓x=5: 25+2=34(neither)x=6: 26+2=66(div by 3) ✓x=7: 27+2=130(neither)\begin{aligned} &x=0:\ 2^0+2=3 \quad\text{(div by 3)}\ \checkmark\\ &x=1:\ 2^1+2=4 \quad\text{(div by 4)}\ \checkmark\\ &x=2:\ 2^2+2=6 \quad\text{(div by 3)}\ \checkmark\\ &x=3:\ 2^3+2=10 \quad\text{(neither)}\\ &x=4:\ 2^4+2=18 \quad\text{(div by 3)}\ \checkmark\\ &x=5:\ 2^5+2=34 \quad\text{(neither)}\\ &x=6:\ 2^6+2=66 \quad\text{(div by 3)}\ \checkmark\\ &x=7:\ 2^7+2=130 \quad\text{(neither)} \end{aligned}
3

Count the successes. The qualifying values are x=0,1,2,4,6x=0,1,2,4,6.

Number of integers=5\text{Number of integers}=5
Count=5\text{Count}=5

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CAT 2018 Slot 1 QA Q27: The number of integers x such that 0.25 &le; 2 x &le; 200, and 2 x + 2 is perfectly divisible by either 3 or 4 — Solution | TheCATExam