whats the answer please??
38
Since the sum of the digits is divisible by 3, the original number is also divisible by 3.Since the sum of the digits is divisible by 3, the original number is also divisible by 3.Since the sum of the digits is divisible by 3, the original number is also divisible by 3.Since the sum of the digits is divisible by 3, the original number is also divisible by 3.
55
Possibility of two digit no whose sum is 9 18,27,36,45,54,63,72,81 Subract 9 with each no mentioned above output is 9,18,27,36,45,54,63,72 See after 4th comma 54 and 45. Reverse 54=45. now 45 is 9 less than 54. So the original no is 54
you just carry the first number and add 5 to the second number
38
Multiply the last digit by 7. Subtract that number from the remaining digits. If that number is divisible by 23, then the original number is divisible by 23.
Reverse the digits then check of the new number is the same as the original number.
Possibility of two digit no whose sum is 17 89 and 98 Reverse of 89 is 98. 98 is 9 less than the original no 89. 89 is original no
Any four-digit number will do. Add it to the number 33333, to get the original number you have to subtract from.Any four-digit number will do. Add it to the number 33333, to get the original number you have to subtract from.Any four-digit number will do. Add it to the number 33333, to get the original number you have to subtract from.Any four-digit number will do. Add it to the number 33333, to get the original number you have to subtract from.
2178 x 4 = 8712
The most efficient method is to reverse the digits in the number and then check if the reversed number is equal to the original number. In C, we can implement this as follows (using long-hand notation for clarity). Note the use of base (defaulting to base 10) so that we can check values in other bases. bool is_palindrome (unsigned num, unsigned base = 10) { // local variables unsigned rev, digit, digits; // copy the input number digits = num; // initialise the accumulator rev = 0; // repeat while there are digits while (digits) { // extract the low-order digit using modulo (%) operator digit = digit % base; // shift the accumulated digits (if any) one position to the left rev = rev * base; // add on the new digit rev = rev + digit; // shift the remaining digits one position to the right digits = digits / base; } // end while // return true if the number and its reverse are equal return num == rev; }
Possibility of two digit no whose sum is 10 19,28,37,46,55,64,73,82,91 Now add 54 to each no mentioned above 73,82,91,100,109,118,127,136,145 See after 1st comma 28 and 82 Reverse of 28 is 82. That no 82 is 54 more than the no 28. So 28 is the original
Since the sum of the digits is divisible by 3, the original number is also divisible by 3.Since the sum of the digits is divisible by 3, the original number is also divisible by 3.Since the sum of the digits is divisible by 3, the original number is also divisible by 3.Since the sum of the digits is divisible by 3, the original number is also divisible by 3.
let a, b, c denote the three digits of the original number, then the three-digit number is 100a+10b+c. The reverse is 100c+10b+a. Subtract: (100a+10b+c)-(100c+10b+a) to get 99(a-c). Since the digits were decreasing, (a-c) is at least 2 and no greater than 9, so the result must be one of 198, 297, 396, 495, 594, 693, 792, or 891. When you add any one of those numbers to the reverse of itself, you get 1089
34.
You do by subtracting one from the previous number and adding 10 to the top number of the digits you are regrouping. Subtract those two digits and you should get your answer.