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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.

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.

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.
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...
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) –...
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) –...
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) –...
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
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
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
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 = 1 and b = 7  71 - 17 = 54 = 9*6 = multiple of 9
17 + 71 = 88 = 11*8 = a multiple of 11 this can be explained by  (10a + b) + (10b + a) = 11a + 11b = 11 (a + b) here a = 1 and b = 7
64 + 46 = 110 = 11*10= a multiple of 11 this can be explained by  (10a + b) + (10b + a) = 11a + 11b = 11 (a + b) here a = 6 and b = 4
39 + 93 = 133 = 11*12 = a multiple of 11 this can be explained by  (10a + b) + (10b + a) = 11a + 11b = 11 (a + b) Here a = 3 and b = 9 
As we know  (10a + b) + (10b + a) = 11a + 11b = 11 (a + b) here a = 2 and b = 7  27 + 72 = 99= 11 *9 =  a multiple of 11
As we know abc = 100 × a + 10 × b + 1 × c casual form of 100 × a + 10 × b + 1 × c = abc.
As we know abc = 100 × a + 10 × b + 1 × c the usual form of number  100*7 + 10*1 + 8 = 700 + 10 + 8 = 718.
As we know ab = 10 × a + b = 10a + b usual form of number 10*5 + 6 = 50 + 6 = 56
A 3-digit number abc made up of digits a, b and c is written as abc = 100 × a + 10 × b + 1 × c = 100a + 10b + c hence generalised form of  302 = 100*3 + 10*0 + 1*2
A 3 digit number madeup of digits a, b, c will be written as, abc = 100 × a + 10 × b + 1 × c = 100a + 10b + c hence generalised form of number 129 = 100*1 + 10*2 + 1*9
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