### 24

 Divisibility by 24 Method 1: If all the digits except the digit at the hundreds place, tens place and unit place are added. The result is multiplied by 16. The digit at the hundreds place is multiplied by 4. The digit at the tens place is multiplied by 10. The results of step 2,3 and 4 are added to the digit at the unit place. If result of step 5 is large then the step 1, 2, 3, 4 and 5 is repeated in order. If the result of 6 is divided by 24 and the remainder is zero then the number is divisible by 24 else not. Method 2:⇒ If the number is divisible by 8 and 3 both then the number is divisible by 24. Proof: Let the number be abcde. abcde can be written as a×10000 + b×1000 + c×100 + d×10 + e which can also be written as a×(24×416+10) + b×(24×55+16) + c×(24×4+4) + d×10 + e Every power of 10 i.e. 10x when divided by 24 gives 16 as remainder,x>2 and is a positive integer. Proof for remainder of 10x when divided by 24 100 = 24×4 + 4 1000 = 10×100 =10×(24×4 + 4) =10×24×4 + 40 =10×24×4 + 24 + 16 remainder = 16 10000 = 10×1000 =10(10×24×4 + 24 + 16) =10(10×24×4 + 24) + 160 160=6×24 + 16 We get 160 in every power of ten with exponent greater than 3, so remainder is 16 every time. So the number is divisible if the number satisfies Method 1. The number is divisible by 24 if [(sum of all digits except the last three digits)×16 + (digit at the hundreds place)×4 + (digit at the tens place)×10 + digit at unit place ] is divisible by 24. Remainder The remainder is the remainder obtained in method 1. Examples Is 340 divisible by 24. 3×4 + 4×10 + 0 = 12+40=52 Remainder when 52 is divided by 24 = 4 The number is not divisible and the remainder is 4. Is 3483 divisible by 24. 3×16+4×4+8×10 + 3 = 48+16+80+3 147 Repeat the process , 1×4+4×10+7 = 4+40+7 = 51 Remainder when 51 is divided by 24 = 3 The number is not divisible and the remainder is 3.