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P Pankaj Sanodiya
Any number will be divisible by 9 only if the sum of all the digits in that number will be divisible by 9. Sum of digits of 927 = 9 + 2 + 7 = 18 which is divisible by 18. Hence we conclude number 927 is divisible by 9.

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P Pankaj Sanodiya
Any number will be divisible by 9 only if the sum of all the digits in that number will be divisible by 9. Sum of digit 432 = 4 + 3 + 2 = 9 which is divisible by 9. Hence we conclude 432 is divisible by 9.

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P Pankaj Sanodiya
Any number will be divisible by 9 only if the sum of all the digits in that number will be divisible by 9. Sum of digits of 294 = 2 + 9 + 4 = 15 which is not divisible by 9. Hence we conclude 294 is not divisible by 9.

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P Pankaj Sanodiya
Any number will be divisible by 9 only if the sum of all the digits in that number will be divisible by 9. Sum of digits of 616 = 6 + 1 + 6 = 13 which is not divisible by 9, Hence we conclude 616 is not divisible by by 9.

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P Pankaj Sanodiya
Any number will be divisible by 9 only if the sum of all the digits in that number will be divisible by 9. Sum of digit of 108 = 1 + 0 + 8 = 9 which is divisible by 9 i.e (9/9 = 1). hence we conclude 108 is divisible by 9.

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P Pankaj Sanodiya
Since N leaves the remainder of 4 when divided by 5. the possible values in ones place of number N are 4 or 9. now, since it leaves a remainder of 1 when divided by 2, the N would be an odd number. hence ones digit of N is also an odd number. which means ones digit of our number N is 9.  

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P Pankaj Sanodiya
The detailed solution for the above-written question is as follows N is Even; so it's unit digit is even. Therefore, the unit digit must be 2, 4, 6, 8 or 0.

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P Pankaj Sanodiya
The detailed solution for the above-written question is as follows If the unit digit of a number is 0 or 5, then it is divisible by 5. hence if we need the remainder of 4 then unit digit of number should be 4 or 9.

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P Pankaj Sanodiya
The detailed solution for the above-written question is as follows If a number is divisible by 5 then it's unit digit must be 0 or 5. so if we need the remainder of 1 when divided by 5 then the numbers unit digit must be 1 or 6.

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P Pankaj Sanodiya
Let choosen number be abc then, abc = 100a + 10b + c cab = 100c + 10a + b bca = 100b + 10c + a The addition of all above three, abc + cab + bca = 111(a + b + c) = 37 × 3(a + b + c), which is divisible by 37 here a = 9, b = 3, and c = 7 937 + 793 + 397 = 2109 = 37*57  i.e. divisible by 37.

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P Pankaj Sanodiya
Let choosen number be abc then, abc = 100a + 10b + c cab = 100c + 10a + b bca = 100b + 10c + a The addition of the above all three, abc + cab + bca = 111(a + b + c) = 37 × 3(a + b + c), which is divisible by 37 here a = 1, b = 1, and c = 7 117 + 711 + 117 = 999 = 37*27  i.e. divisible by 37.

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P Pankaj Sanodiya

Let choosen number be abc then,

abc = 100a + 10b + c

cab = 100c + 10a + b

bca = 100b + 10c + a

abc + cab + bca = 111(a + b + c) = 37 × 3(a + b + c), which is divisible by 37

here a = 6, b = 3, and c = 2

632+ 263+ 362= 1221= 37*33 i.e. divisible by 37.

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P Pankaj Sanodiya
Let choosen number be abc then, abc = 100a + 10b + c cab = 100c + 10a + b bca = 100b + 10c + a After adding all the above three, abc + cab + bca = 111(a + b + c) = 37 × 3(a + b + c), It will be divisible by 37 becuase 37 is present in the equation. here a = 4, b = 1, and c = 7 417 + 741 + 147 = 1332 = 37*36  i.e. divisible by 37.

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P Pankaj Sanodiya
Let the 3-digit number chosen by Minakshi = 100a + 10b + c. After reversing the order of the digits, number = 100c + 10b + a. On subtraction: • If a > c, then the difference between the original number & reversed number is (100a + 10b + c) – (100c + 10b + a) = 100a + 10b + c – 100c – 10b – a = 99a – 99c = 99(a – c). • If c > a, then the difference between the numbers is (100c + 10b + a) – (100a...

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P Pankaj Sanodiya
Let the 3-digit number chosen by Minakshi = 100a + 10b + c.  After reversing the order of the digits, number = 100c + 10b + a. On subtraction: • If a > c, then the difference between the original number & reversed number is (100a + 10b + c) – (100c + 10b + a) = 100a + 10b + c – 100c – 10b – a = 99a – 99c = 99(a – c). • If c > a, then the difference between the numbers is (100c + 10b + a) –...

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P Pankaj Sanodiya
Let the 3-digit number chosen by Minakshi = 100a + 10b + c. After reversing the order of the digits, number = 100c + 10b + a. On subtraction: • If a > c, then the difference between the original numbedr & reversed numbers is (100a + 10b + c) – (100c + 10b + a) = 100a + 10b + c – 100c – 10b – a = 99a – 99c = 99(a – c). • If c > a, then the difference between the numbers is (100c + 10b + a) –...

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P Pankaj Sanodiya
Let's assume the 3-digit number chosen by Minakshi = 100a + 10b + c. After reversing the order of the digits, number = 100c + 10b + a. On subtraction: • If a > c, then the difference between the original number & reversed number (100a + 10b + c) – (100c + 10b + a) = 100a + 10b + c – 100c – 10b – a = 99a – 99c = 99(a – c). • If c > a, then the difference between the numbers is (100c + 10b + a) –...

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P Pankaj Sanodiya
If the tens digit is larger than the ones digit (that is, a > b), then (10a + b) – (10b + a) = 10a + b – 10b – a = 9a – 9b = 9(a – b). If the unit digit is larger than the tens digit (that is, b > a), he does: (10b + a) – (10a + b) = 9(b – a). here a = 3 and b = 7  73 - 37 = 36 = 9*4 = multiple of 9

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P Pankaj Sanodiya
If the tens digit is larger than the ones digit (that is, a > b), then (10a + b) – (10b + a) = 10a + b – 10b – a = 9a – 9b = 9(a – b). If the unit digit is larger than the tens digit (that is, b > a), he does: (10b + a) – (10a + b) = 9(b – a). here a = 9 and b = 6 96 - 69 = 27= 9*3= multiple of 9

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P Pankaj Sanodiya
If the tens digit is larger than the ones digit (that is, a > b), then (10a + b) – (10b + a) = 10a + b – 10b – a = 9a – 9b = 9(a – b). If the unit digit is larger than the tens digit (that is, b > a), he does: (10b + a) – (10a + b) = 9(b – a). here a = 2 and b = 1 21 - 12 = 9 = 9*1 = multiple of 9
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