Math IOQM

Q1. A grocer makes a display of cans in which the top row has one can and each lower row has two more cans than the row above it. If the display contains 100 cans, how many rows are there? Solution: 1 + 3 + .. (2n-1) = 100 = n/2[2.1 + (n-1).2] => 100 = n.[1 + n - 1] = n^2 => n = 10.  Q2. The second and fourth terms of a geometric sequence are 2 and 6. Find the possible first terms. Solution: r^2 = 3, ar = 2 => a = -+2/sqrt(3) Q3.  Figures 0, 1, 2, and 3 consists of 1, 5, 13, and 25 nonoverlapping unit squares, respectively. If the pattern were continued, how many non-overlapping unit squares would there be in Figure 100? Solution: a1 = 1 + 4 a2 = 1 + 4 + 8 a3 = 1 + 4 + 8 + 12 a100 = 1 + (4 + 8 + 12 .. 100 times) = 1 + 100/2[8 + 99.4] = 1 + 200.[2 + 99] = 1 + 101.200 = 20201 Q4.  Let 1, 4, . . . and 9, 16, . . . be two arithmetic sequences. The set S is the union of the first 2004 terms of each sequence. How many distinct numbers are in S? Solution: Total terms ...

19. Let N = 6 + 66 + 666 + ⋯ + 666 … 66 N = 6 + 66 + 666 + \dots + 666\ldots66 , where there are hundred 6’s in the last term in the sum. How many times does the digit 7 occur in the number N N ? [PRE-RMO – 2018] Solution N = 6 + 66 + 666 + ⋯ + 666 … 6 ⏟ 100  times N = 6 + 66 + 666 + \dots + \underbrace{666\ldots6}_{100\text{ times}} N = 6 9 [ ( 10 − 1 ) + ( 10 2 − 1 ) + ⋯ + ( 10 100 − 1 ) ] = 6 9 [ 10 9 ( 10 100 − 1 ) − 100 ] = 2 3 [ 999 … 90 ⏟ (100 digits of 9, ending in 0) 9 − 100 ] = 2 3 [ 111 … 110 ⏟ (100 digits of 1, ending in 0) − 100 ] = 2 3 × 111 … 1010 ⏟ (some rearrangement, 98 digits total) = 1 3 × 222 … 2020 ⏟ (98 digits total) = 740740 … 7407340 ⏟ (96 digits total) . \begin{aligned} N &= \frac{6}{9}\Bigl[(10-1) + (10^2 - 1) + \dots + (10^{100} - 1)\Bigr] \\[6pt] &= \frac{6}{9}\Bigl[\frac{10}{9}\bigl(10^{100} - 1\bigr) - 100\Bigr] \\[6pt] &= \frac{2}{3}\...

16. What is the value of ∑ 1 ≤ i< j ≤ 10 i + j = odd ( i + j ) − ∑ 1 ≤ i< j ≤ 10 i + j = even ( i + j ) ? \sum_{\substack{1 \le i \le j \le 10 \\ i + j = \text{odd}}} (i + j) - \sum_{\substack{1 \le i \le j \le 10 \\ i + j = \text{even}}} (i + j)? [PRE-RMO – 2018] To compute the sum when i +j is odd, let's figure out how many times do we need to add something to 10: 10 will be added to 1,3,5,7,9 9 will be added to 2,4,6,8 (don't redo 10 since it's already done above) 8 => 1,3,5,7 (Just note the numbers smaller than this since the bigger ones are already added above). 7 => 2,4,6 6 => 1,3,5 5 => 2,4 4 => 1,3 3 => 2 2 => 1 How many times does each number occur? Each number occurs 5 times. Because each odd number will be added to each of 5 even numbers. And each even number will be added to each of 5 odd numbers. Now to compute the sum when i + j is even: 10 will be added to 2,4,6,8 9 => 1,3,5,7 8 => 2,4,6 7 => 1,3,5 6 => 2,4 5 =...

15. Integers 1, 2, 3, ..., n where n > 2 are written on a board. Two numbers m , k (such that 1 < m < n , 1 < k < n ) are removed and the average of the remaining numbers is found to be 17. What is the maximum sum of the two removed numbers? [PRE-RMO – 2017] Solution: Let two numbers moved be 1, 2. Then average of remaining (n − 2) numbers x ˉ 1 = n ( n + 1 ) 2 − 3 ÷ ( n − 2 ) = n 2 + n − 6 2 ( n − 2 ) = ( n + 3 ) ( n − 2 ) 2 ( n − 2 ) = n + 3 2 \bar{x}_1 = \frac{n(n+1)}{2} - 3 \div (n - 2) = \frac{n^2 + n - 6}{2(n - 2)} = \frac{(n + 3)(n - 2)}{2(n - 2)} = \frac{n + 3}{2} If two numbers be n , n − 1 n, n - 1 , then x ˉ 2 = n ( n + 1 ) 2 − ( 2 n − 1 ) ÷ ( n − 2 ) = n 2 − 3 n + 2 2 ( n − 2 ) = n − 1 2 \bar{x}_2 = \frac{n(n + 1)}{2} - (2n - 1) \div (n - 2) = \frac{n^2 - 3n + 2}{2(n - 2)} = \frac{n - 1}{2} Now, the removed numbers are m , k m, k , then new average is 17 i.e., 17 = n ( n + 1 ) 2 − ( m + k ) ÷ ( n − 2 ) 17 = \frac{n(n + 1)}{2} - (m + k)...