Five.
The five numbers are 11, 12, 15, 24, 36
If you mean, "What is the largest number of digits possible in the product of two 2-digit numbers" then 99 * 99 = 9801, or 4 digits. Anything down to 59 * 17 = 1003 will have 4 digits.
You had me until "product." The product of 4 digits can't be prime.
It can have 4 digits, because the highest possible two digit numbers 99*99=9801.
The product of a 4-digit multiplicand and a 1-digit multiplier can have either 4 or 5 digits. If the 4-digit number is multiplied by a multiplier of 1 to 9, the product will typically have 4 digits. However, if the multiplicand is multiplied by 10 (the maximum value for a 1-digit number), the product can reach a maximum of 5 digits. Thus, the product can range from 4 to 5 digits, depending on the specific numbers involved.
There are 15 of them.
from 3 digits (10x10) to 4 digits (99X99)
To find the last but one digit in the product of the first 75 even natural numbers, we need to consider the units digit of each number. Since we are multiplying even numbers, the product will end in 0. Therefore, the last but one digit (tens digit) will depend on the multiplication of the tens digits of the numbers. The tens digit will be determined by the pattern of the tens digits of the even numbers being multiplied.
-21
There are three such numbers: 12, 24 and 36.
4 Give examples to support your answer please.
The number of zeros in the product of multi-digit numbers with zeros and one-digit numbers depends on the placement of the zeros in the multi-digit numbers. If a zero is at the end of a multi-digit number, it effectively multiplies the other digits by ten, contributing to the count of zeros in the product. However, if the zeros are located elsewhere, they may not affect the overall count of zeros in the final product. Thus, the final count of zeros can vary based on the specific arrangement of digits.
There are five such numbers: 11, 12, 15, 24 and 36.