Want this question answered?
Be notified when an answer is posted
The 2-digit number must be 20, because it is the only 2-digit number whose sum of its two even digits, 2 + 0 = 2, is greater than the product of its two even digits, 2 x 0 = 0. Moreover, 20 is a product of the two consecutive integers 4 and 5.
There are seven digits: 8,000,000
231
225 = (15)2 2*2*5=20 3 digits in 225
81
You could be 22? 2 + 2 = 4 2 x 2 = 4
The 2-digit number must be 20, because it is the only 2-digit number whose sum of its two even digits, 2 + 0 = 2, is greater than the product of its two even digits, 2 x 0 = 0. Moreover, 20 is a product of the two consecutive integers 4 and 5.
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.
36 is twice the product of its digits multiplied. 3 times 6 equals 18, and 18 times 2 equals 36.
from 3 digits (10x10) to 4 digits (99X99)
19
There are seven digits: 8,000,000
231
2 digits because 1.88*71 = 133.48
225 = (15)2 2*2*5=20 3 digits in 225
20
81