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CAT 2021 Slot 3 — QA Question 11

IndicesEasy

In ‘n’ is a positive integer such that (107)(107)2...(107)n > 999, then the smallest value of n is

Answer & solution

Correct answer: 6

Answer: 6

Solution

Easy

The product is (107)1(107)2⋯(107)n=10(1+2+⋯+n)/7\big(\sqrt[7]{10}\big)^{1}\big(\sqrt[7]{10}\big)^{2}\cdots\big(\sqrt[7]{10}\big)^{n}=10^{(1+2+\cdots+n)/7}. Add the exponents using the sum 1+2+⋯+n=n(n+1)21+2+\cdots+n=\tfrac{n(n+1)}{2}, then compare the power of 1010 with 999999 (just under 10310^3).

1

Combine into a single power of 1010. Each factor is 10k/710^{k/7}; multiplying adds exponents.

∏k=1n10k/7=10(1+2+⋯+n)/7⇒ =10 n(n+1)/14(∑k=n(n+1)2)\begin{aligned} &\prod_{k=1}^{n}10^{k/7}=10^{(1+2+\cdots+n)/7}\\ &\Rightarrow\ =10^{\,n(n+1)/14} \quad\text{(}\textstyle\sum k=\tfrac{n(n+1)}{2}\text{)} \end{aligned}
2

Turn the inequality into a condition on nn. Since 999<1000=103999<1000=10^3, the cleanest sufficient threshold is reaching 10310^3, i.e. exponent ≥3\ge 3.

10 n(n+1)/14>999⇐ n(n+1)14≥3(then 10n(n+1)/14≥1000>999)⇒ n(n+1)≥42\begin{aligned} &10^{\,n(n+1)/14}>999\\ &\Leftarrow\ \frac{n(n+1)}{14}\ge 3 \quad\text{(then }10^{n(n+1)/14}\ge 1000>999\text{)}\\ &\Rightarrow\ n(n+1)\ge 42 \end{aligned}
3

Smallest valid nn. Test consecutive integers for n(n+1)≥42n(n+1)\ge 42.

n=5: 5⋅6=30<42(too small)n=6: 6⋅7=42≥42(works)\begin{aligned} &n=5:\ 5\cdot 6=30<42 \quad\text{(too small)}\\ &n=6:\ 6\cdot 7=42\ge 42 \quad\text{(works)} \end{aligned}
Smallest n=6\text{Smallest }n=6

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CAT 2021 Slot 3 QA Q11 — Indices solution | TheCATExam