Question 10 of 17advanced🔍 AnalyzeLong Answer4 marks

Consider any 3-digit number, say abc (100a + 10b + c). Make two other 3-digit numbers from these digits by cycling these digits around, yielding bca and cab. Now add the three numbers. Using algebra, justify that the sum is always divisible by 37. Will it also always be divisible by 3?

Correct Answer

Step 1: Write abc = 100a + 10b + c

Step 2: Write bca = 100b + 10c + a

Step 3: Write cab = 100c + 10a + b

Step 4: Sum = (100a + 10b + c) + (100b + 10c + a) + (100c + 10a + b)

Step 5: Sum = 111a + 111b + 111c = 111(a + b + c)

Step 6: Since 111 = 3×373 \times 37, the sum is divisible by both 37 and 3.

Answer: The sum is always divisible by both 37 and 3.

Exercise: Figure it Out (Section 3) | Q: 3 | (Page Page 11)
For More Understanding

Explanation

The chapter encourages using algebra to justify mathematical claims. By expressing each cyclic permutation algebraically and adding them, we get 111(a + b + c). Since 111 = 3×373 \times 37, the sum is divisible by both 37 and 3.

Solution Steps

  1. Step 1: Write abc = 100a + 10b + c

  2. Step 2: Write bca = 100b + 10c + a

  3. Step 3: Write cab = 100c + 10a + b

  4. Step 4: Sum = (100a + 10b + c) + (100b + 10c + a) + (100c + 10a + b)

  5. Step 5: = 111a + 111b + 111c = 111(a + b + c)

  6. Step 6: Since 111 = 3×373 \times 37, the sum is divisible by both 37 and 3

PreviousNext