Question 3 - Figure it out - Page 145-147 - Chapter 6 Class 8 - Algebra Play (Ganita Prakash II)

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Question 3 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? [Hint: Look at some multiples of 37.]Writing the 3-digit number in place value form abc = 100a + 10b + c bca = 100b + 10c + a cab = 100c + 10a + b Adding all there numbers Sum = (100a + 10b + c) + (100b + 10c + a) + (100c + 10a + b) = (100a + 10a + a) + (100b + 10b + b) + (100c + 10c + c) = 111a + 111b + 111c = 111 (a + b + c) = 37 × 3 × (a + b + c) Since sum is 37 is multiplied to a number 3 × (a + b + c) Therefore, if you divide it by 37, it will always divide perfectly with a remainder of zero. And, since 3 is also multiplied to the number We can also that, if you divide it by 3, it will always divide perfectly with a remainder of zero.