2178 x 4 = 8712
whats the answer please??
38
When multiplying numbers with significant digits, count the total number of significant digits in each number. Multiply the numbers as usual, but round the final answer to match the least number of significant digits in the original numbers.
Reverse the digits then check of the new number is the same as the original number.
To total 17 the two digits must be 8 and 9! The original number was 98.
The 11 finger trick is a math trick where you can quickly multiply numbers by 11. To use the trick, you separate the digits of the number you want to multiply by 11, add the two digits together, and place the result in between the original digits. This works because 11 is the same as 10 1, so when you add the two digits together and place the result in between, you are essentially multiplying the original number by 10 and then adding the original number to it.
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.
five digits number when you multiply by four is the same the number when you reversed it is 21978*4 = 87912
the number one.
The 11 finger trick is a mental math technique where you can quickly multiply numbers by 11. To use this trick, you separate the digits of the number you want to multiply by 11, add the digits together, and place the sum in the middle of the original number. This allows you to calculate the product of the original number and 11 without using traditional multiplication methods.
by moving each digits
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; }