HOTS Question Answers: Number Play

Q1: Kiran has an odd number of ₹2 coins and an even number of ₹5 coins. Can the total amount be ₹37? Explain why.

Ans: No
Explanation: Each ₹2 coin contributes an even amount because 2 × (odd number) = even.
Each ₹5 coin contributes an even amount because (even number) × 5 = even.
Their sum is even (even + even = even). 
Since ₹37 is odd, it is impossible to obtain ₹37 by adding two even amounts.
Example: 3 two-rupee coins give 6 and 4 five-rupee coins give 20; 6 + 20 = 26, which is even.

Q2: Create an algebraic expression that always gives odd numbers for any integer n.

Ans: 2n + 1
Explanation: For any integer n, 2n is always even because it is a multiple of 2. Adding 1 to an even number makes it odd.
Examples: n = 1 → 2×1 + 1 = 3; n = 0 → 2×0 + 1 = 1; n = 2 → 2×2 + 1 = 5.

Q3: In a 3 × 3 magic square using numbers 1 to 9, why can't the number 8 be in the center?

Ans:  8 cannot be in the center.

Explanation:
The magic sum is 15.
If 8 is in the center, the other two numbers in each line must add to
$15-8=715\; -\; 8\; =\; 7$15-8=7.

The pairs that add to 7 are:
1 + 6, 2 + 5, and 3 + 4.

There are only three such pairs, but the center needs four pairs.
So 8 cannot be placed in the center.

Q4: If each number in a 3 × 3 magic square (numbers 1 to 9) is multiplied by 3, is the result still a magic square? What is the new magic sum?

Ans: Yes, the new magic  sum is 45
Explanation: The original magic sum is 15.
Multiplying every number by 3 also multiplies the magic sum by 3.
So the new magic sum is $15\times 3=4515\; \backslash times\; 3\; =\; 45$
The square is still a magic square.

Q5: Maya climbs a 7-step staircase, taking 1 or 2 steps at a time. How many ways can she do this?

Ans: 21
Explanation: Let f(n) be the number of ways to climb n steps with 1- or 2-step moves. For n ≥ 3, each way either starts with a single step (leaving n-1 steps) or a double step (leaving n-2 steps). So f(n) = f(n-1) + f(n-2).
Base values: f(1) = 1, f(2) = 2.
Compute: f(3) = 3, f(4) = 5, f(5) = 8, f(6) = 13, f(7) = 21.
Thus Maya can climb 7 steps in 21 different ways.

Q6: Design a 2 × 3 grid with 3 odd and 3 even numbers so that all row sums are odd and all column sums are even. Is this possible?

Ans: No
Explanation: Let the grid have 2 rows and 3 columns (total 6 entries) and exactly 3 odd numbers overall. For a row of three numbers to have an odd sum, that row must contain an odd number of odd entries (either 1 or 3 odd entries). Since there are two rows, the total number of odd entries required is the sum of two odd numbers, which is always even (for example 1+1=2 or 1+3=4 or 3+3=6). But the grid is specified to have exactly 3 odd numbers in total, which is odd. This is a contradiction. Therefore it is impossible to arrange 3 odd and 3 even numbers in a 2 × 3 grid so that every row sum is odd and every column sum is even.

Q7: In a 3 × 3 magic square with center number 10, what is the magic sum?

Ans: 30
Explanation:

In a 3 × 3 magic square, the magic sum is always three times the center number.
So,

$Magic sum=3\times 10=30\backslash text\{Magic\; sum\}\; =\; 3\; \backslash times\; 10\; =\; 30$Magic sum=3×10=30

Q8: Arjun has 60 loose sheets from a book, each printed on both sides. Can the sum of the page numbers be 7000? Why or why not?

Ans: No
Explanation: Each sheet has two consecutive page numbers, so 60 sheets give 120 consecutive pages. Let the first page number be n; then the pages are n, n+1, ..., n+119.
The sum of these 120 pages is 120n + (0 + 1 + ... + 119) = 120n + (119×120)/2 = 120n + 7,140.
For this sum to be 7,000 we would need 120n + 7,140 = 7,000, which gives 120n = -140, impossible for positive n.
Hence no set of 120 consecutive page numbers from a book can sum to 7,000.