Figure it Out
Figure it Out - Solutions
1. The sum of four consecutive numbers is 34. What are these numbers?
Let the four consecutive numbers be .
Their sum is:
The four numbers are: 7, 8, 9, 10
2. Suppose p is the greatest of five consecutive numbers. Describe the other four numbers in terms of p.
If p is the greatest, the five consecutive numbers in increasing order are:
The other four numbers are: p−4, p−3, p−2, and p−1
3. Determine if each statement is always true, sometimes true, or never true.
(i) The sum of two even numbers is a multiple of 3.
SOMETIMES TRUE
- Example where true: 3 + 6 = 9 (multiple of 3) ✓
- Example where false: 2 + 4 = 6... wait, that's a multiple of 3
- Counter-example: 2 + 2 = 4 (not a multiple of 3) ✓
Algebraically: Let the two even numbers be 2a and 2b.
This is always even, but whether it's divisible by 3 depends on whether is divisible by 3. This is not guaranteed.
(ii) If a number is not divisible by 18, then it is also not divisible by 9.
NEVER TRUE
Counter-example: 9 is not divisible by 18, but 9 IS divisible by 9. ✓
Also: 27, 45, 63, etc. are all divisible by 9 but not 18.
Algebraically: If n is divisible by 9, then n = 9k for some integer k. For n to also be divisible by 18, we need n = 18m, which means 9k = 18m, or k = 2m. This is true only when k is even, not always.
(iii) If two numbers are not divisible by 6, then their sum is not divisible by 6.
SOMETIMES TRUE
- Counter-example: 3 and 9 are both not divisible by 6, but 3 + 9 = 12, which is not divisible by 6. ✓
- Counter-example showing it can be divisible: 2 and 4 are not divisible by 6, but 2 + 4 = 6, which IS divisible by 6. ✗
The statement is false in some cases.
(iv) The sum of a multiple of 6 and a multiple of 9 is a multiple of 3.
ALWAYS TRUE
Algebraically:
- Multiple of 6:
- Multiple of 9:
- Sum:
This is always a multiple of 3. ✓
(v) The sum of a multiple of 6 and a multiple of 3 is a multiple of 9.
SOMETIMES TRUE
- Example where true: 6 + 12 = 18 (multiple of 9) ✓
- Counter-example: 6 + 3 = 9... actually that is a multiple of 9
- Counter-example: 6 + 6 = 12 (not a multiple of 9) ✓
Algebraically:
- Multiple of 6:
- Multiple of 3:
- Sum:
This is a multiple of 3, but not necessarily 9. For it to be a multiple of 9, we'd need to be a multiple of 3, which is not always true.
4. Find numbers that leave remainder 2 when divided by 3 and by 4.
Numbers leaving remainder 2 when divided by 3: for integer k
Numbers leaving remainder 2 when divided by 4: for integer m
We need both conditions:
Since gcd(3,4) = 1, we need k = 4t and m = 3t.
So:
Examples: 2, 14, 26, 38, 50, ...
General form: n = 12t + 2, or equivalently, n ≡ 2 (mod 12)
Or: n = 12n + 2 where n is a non-negative integer
5. The Pebbles Puzzle
We need a number n where:
- n ≡ 1 (mod 3)
- n ≡ 1 (mod 2), i.e., n is odd
- n ≡ 1 (mod 5)
- n ≡ 0 (mod 7)
- 0 < n ≤ 100
From the first three conditions, by the Chinese Remainder Theorem:
(since lcm(3,2,5) = 30)
So n = 30j + 1 for some integer j.
From the fourth condition:
Testing values:
- j = 1: 31 ≡ 3 (mod 7)
- j = 2: 61 ≡ 5 (mod 7)
- j = 3: 91 ≡ 0 (mod 7) ✓
The number of pebbles is 91.
Verification: 91 ÷ 3 = 30 remainder 1 ✓, 91 is odd ✓, 91 ÷ 5 = 18 remainder 1 ✓, 91 ÷ 7 = 13 remainder 0 ✓
6. Tathagat's Claim
Numbers leaving remainder 2 when divided by 6:
Sum of three such numbers:
This is always a multiple of 6.
Tathagat's claim is TRUE. ✓
7. Remainders when divided by 7
Given:
- 661 ≡ 3 (mod 7)
- 4779 ≡ 5 (mod 7)
(i) 4779 + 661
Remainder: 1
Visual representation: If we think of remainders on a circle (0-6), moving 5 steps and then 3 more steps gives us 8, which wraps around to 1.
(ii) 4779 − 661
Remainder: 2
8. Find the smallest number with specific remainders
We need:
- n ≡ 2 (mod 3)
- n ≡ 3 (mod 4)
- n ≡ 4 (mod 5)
Notice the pattern: Each remainder is one less than the divisor!
- n ≡ -1 (mod 3)
- n ≡ -1 (mod 4)
- n ≡ -1 (mod 5)
This means:
So is a common multiple of 3, 4, and 5.
Therefore:
The smallest positive value:
The smallest such number is 59.
Verification: 59 ÷ 3 = 19 remainder 2 ✓, 59 ÷ 4 = 14 remainder 3 ✓, 59 ÷ 5 = 11 remainder 4 ✓
Why it's the smallest: The pattern (remainder = divisor − 1) means n + 1 must be divisible by all three numbers, making lcm(3,4,5) = 60 the key. The smallest positive n is therefore 59.
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