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CAT 2024 Slot 1 — QA Question 4

Number TheoryEasy

For any natural number n, let an be the largest integer not exceeding √n. Then the value of a1 + a2 + a3 + ... + a50 is

Answer & solution

Correct answer: 217

Answer: 217

Solution

Medium

an=⌊n⌋a_n=\lfloor\sqrt n\rfloor stays constant on each band between consecutive perfect squares. Count how many nn fall in each band, multiply by the constant value, and add the leftover tail up to 5050.

1

Identify the bands. For k2≤n≤(k+1)2−1k^2\le n\le (k+1)^2-1 we have an=ka_n=k. Each full band has (k+1)2−k2=2k+1(k+1)^2-k^2=2k+1 values.

k=1: n=1…3  (3 values)⇒ k=2: n=4…8  (5 values)⇒ k=3: n=9…15  (7 values)⇒ k=4: n=16…24  (9 values)⇒ k=5: n=25…35  (11 values)⇒ k=6: n=36…48  (13 values)\begin{aligned} &k=1:\ n=1\ldots3\ \ (3\text{ values})\\ &\Rightarrow\ k=2:\ n=4\ldots8\ \ (5\text{ values})\\ &\Rightarrow\ k=3:\ n=9\ldots15\ \ (7\text{ values})\\ &\Rightarrow\ k=4:\ n=16\ldots24\ \ (9\text{ values})\\ &\Rightarrow\ k=5:\ n=25\ldots35\ \ (11\text{ values})\\ &\Rightarrow\ k=6:\ n=36\ldots48\ \ (13\text{ values}) \end{aligned}
2

Handle the tail. The next band k=7k=7 would start at 4949, and since 49,50≤5049,50\le 50, we get a49=a50=7a_{49}=a_{50}=7 (22 values).

n=49,50: an=7  (2 values)\begin{aligned} &n=49,50:\ a_n=7\ \ (2\text{ values}) \end{aligned}
3

Sum value ×\times count over all bands from steps 1 and 2.

S=1(3)+2(5)+3(7)+4(9)+5(11)+6(13)+7(2)⇒ S=3+10+21+36+55+78+14(multiply each)⇒ S=217\begin{aligned} &S=1(3)+2(5)+3(7)+4(9)+5(11)+6(13)+7(2)\\ &\Rightarrow\ S=3+10+21+36+55+78+14 \quad\text{(multiply each)}\\ &\Rightarrow\ S=217 \end{aligned}
a1+a2+⋯+a50=217a_1+a_2+\cdots+a_{50}=217

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CAT 2024 Slot 1 QA Q4: For any natural number n, let a n be the largest integer not exceeding √n. Then the value of a 1 + a 2 + — Solution | TheCATExam