If there are two digits than the first one is ten times its value.
So as an algebraic equation we get
10A + B - 3 = 7(A x B)
This can be simplified to
10A + B - 3 = 7A + 7B
and this reduces to
10A - 7A = 7B - B +3
or
3A =6B +3
A solution to this is
3(3) = 6(2) +3
Which would make the original equation
32 = 7(3+2)-3
27
26.
804
27
To find the number of three-digit combinations, we consider the digits from 000 to 999. Each digit can range from 0 to 9, giving us 10 options for each of the three digits. Therefore, the total number of three-digit combinations is (10 \times 10 \times 10 = 1,000).
The sum is 22 times the sum of the three digits.
To form a three-digit number using the digits 0-9, the first digit cannot be 0 (as it would not be a three-digit number). Thus, the first digit can be any of the digits from 1 to 9 (9 options). The second and third digits can each be any digit from 0 to 9 (10 options each). Therefore, the total number of three-digit numbers is (9 \times 10 \times 10 = 900).
27
If one can repeat digits, each of the three positions in a three-digit password can be filled by any of the 10 digits (0-9). Therefore, the total number of three-digit numerical passwords is (10 \times 10 \times 10 = 1,000). Thus, there are 1,000 possible three-digit numerical passwords when digits can be repeated.
To form a three-digit number using the digits 1-7, we can choose any of the 7 digits for each of the three places (hundreds, tens, and units). Therefore, the total number of 3-digit combinations can be calculated as (7 \times 7 \times 7), which equals 343. Thus, there are 343 different three-digit numbers that can be formed using the digits 1-7.
it is 10
38977 is in ones place9 is in tens place (and is three times the number in thousands place)8 is in the hundreds place3 is in the thousands place7+9+8+3=27