For the number 25, the 5 is the ones digit (and the 2 is the tens digit). So, for 99 the second 9 from left to right is the ones digit.
The 9 in 879 represents nine ones or nine units
The units digit of 9n is 9 if n is odd and 1 if n is even. So 1.
The units digit of 159*445*7762*39 is the units digit of the product of the units digits of the four numbers, that is, the units digit of 9*5*2*9 Since there is a 5 and a 2 in that, the units digit is 0.
35 and 119.
9
The answer depends on what the tens digit is greater than, and what the ones digit does then.
It is a 3. Look at the ones digit of successive powers of 7; this need only be done by considering the multiplication of the ones digit of the previous power of 7 by 7 (as this is the only calculation that affects the ones digit as each successive power of 7 is the previous power multiplied by 7) and taking the result modulus 10 (to extract the new ones digit as any excess over 9 is carried into the tens column): 7¹ → 7 mod 10 = 7 7² → (7×7) mod 10 = 9 7³ → (9×7) mod 10 = 3 7⁴ → (3×7) mod 10 = 1 7⁵ → (1×7) mod 10 = 7 At this point the pattern of the ones digit will obviously repeat the sequence of the four digits {7, 9, 3, 1}. To find the ones digit of any power of 7, take that power modulus 4 use that digit from the four digit sequence. Note that when taking the number modulus 4, the result will be in the range 0-3; when the result is 0, use the 4th digit from the sequence. 2015 mod 4 = 3 → the third digit of {7, 9, 3, 1}, which is 3, will be the ones digit of 7²⁰¹⁵.
For the number 25, the 5 is the ones digit (and the 2 is the tens digit). So, for 99 the second 9 from left to right is the ones digit.
91 since the ones digit is 8 less than 9 and the two digits, 9+1 = 10, a two-digit number.
9
The 9 in 879 represents nine ones or nine units
If the first digit is 9, you have 9 options (0-8) for the second digit. If the first digit is 8, you have 8 options (0-7) for the second digit. Etc. This leaves you with the arithmetic series: 0 + 1 + 2 + 3 + ... + 9.
Power 2: units digit 9. Multiply by 49 again to get power 4: units digit 1. So every 4th power gives units digit 1. So 16th power has units digit 1, so the previous power, the 15th must have units digit 3.
The composite number 2007 is deficient. Prime factors: 3 * 3 * 223 Prime factors in exponent form: 3^2 * 223 All factors: 1, 3, 9, 223, 669, 2007
If the first digit is 9, you have 9 options (0-8) for the second digit. If the first digit is 8, you have 8 options (0-7) for the second digit. Etc. This leaves you with the arithmetic series: 0 + 1 + 2 + 3 + ... + 9.
The units digit of 9n is 9 if n is odd and 1 if n is even. So 1.