What is the result of the sum of 45 and the quotient divided by 4 / 5?

What is the result of the sum of 45 and the quotient divided by 4 / 5?

eighty-one

The sum of four numbers is 100, the first number plus 4, the second number minus 4, the third number multiplied by 4, and the fourth number divided by 4
The sum of four numbers is 100, the first number plus 4, the second number minus 4, the third number multiplied by 4, and the fourth number divided by 4. Their sum remains unchanged. How many are they?

[(X+4)+(X-4)+(X*4)+(X/4)]=100
4[(X+4+X-4+4X+X/4)=100]
4X+16+4X-16+16X+X=400
25X=400
X=400/25
X=16
1.X+4=20
16+4=20
2.X-4=12
16-4=12
3.X*4=64
16*4=64
4.X\4=4
16/4=4

2. Given two natural numbers a and B (a > b), the remainder of a and B divided by 13 is 5 and 9 respectively. Find the remainder of a + B, A-B, a × B, A2-B2 divided by 13 respectively
3. 2100 divided by a two digit number, the remainder is 56
4. The sum of divisor, divisor, quotient and remainder is 903, the known divisor is 35, and the remainder is 2
5. Remove 345 and 543 with an integer, the remainder is the same, and the quotient difference is 9
6. There is an integer, and the sum of the three remainders obtained by removing 312231123 is 41

2. If a is divided by 13 remainder 5 and B is divided by 13 remainder 9, then: ① a + B is divided by 13 remainder 5 + 9 = 14, 14 pairs of 13 remainder 1, so a + B pairs 13 remainder 1; ② A-B, because a > b, so A-B is divided by 13 remainder (13 + 5) - 9 = 9, borrow forward; ③ a × B, let a = 13m + 5, B = 13N + 9 (M = 1,2,3,..., n = 0,1,2,...), then a × B = (13m + 5) (13N + 9) = 13

2. Find the remainder of 123.1920 divided by 7
Such as the title

1. Mod (2,13) = 2 mod (22,13) = 9mod (222,13) = 1mod (2222,13) = 12mod (22222,13) = 5mod (222222,13) = 0so mod (2010,6) = 0 so mod (2222.2) = 02.1234567891011121314.1920, I use Excel to calculate 7, the remaining 5 formula is as follows: first set 1920 column A1 ~ A10 as 1,2,3

The sum of an integer divided by 139123,48 is 23, and the number of times is?
Please give the answer and find the solution

139 + 123 + 48 - 23 = 287 - after subtracting the remainder from the sum, the difference can be divided by this number
287 = 7 × 41, the original number is 7, because the divisor is 7, the maximum remainder is 6, 6 × 3 = 18

14. There is an integer, and the sum of the three remainders obtained by removing 312231123 is 41

321 / x + 231 / x + 123 / x = 666 / x = KX + 41 K is the integral part of quotient
KX = 666-41 = 525
Because: K and X are integers, KX = 5 * 125 = 25 * 25,
X = 125
So: x = 25
Brought in: 312 / 5 over 12 231 / 5 over 6 123 / 5 over 23
The sum of three remainder: 12 + 6 + 23 = 41

There is a natural number, which can be used to remove 226 odd a, 411 odd a + 1 and 527 odd a + 2, then a=______ ．

410-226 = 184 = 23 × 8525-226 = 299 = 23 × 13525-410 = 115 = 23 × 5 19, that is: a = 19; so the answer is: 19

A natural number, with 226 removed, the remainder is a, 411 removed, the remainder is [a + 1], 527 removed, the remainder is [a + 2], what is a
thinking

411*2-226-527=69
The remainder is 19
The remainder is 66
69 minus 527, the remainder is 44
Therefore, the number is not 69, but 23 (3 is smaller)

There is a natural number, which can be used to remove 226 odd a, 411 odd a + 1 and 527 odd a + 2, then a=______ ．

410-226 = 184 = 23 × 8525-226 = 299 = 23 × 13525-410 = 115 = 23 × 5 19, that is: a = 19; so the answer is: 19

The remainder of a natural number divided by 326, 258 and 207 is the same. What is the natural number?

326-258 = 68258-207 = 51, 68 = 17 × 4, 51 = 17 × 3, so the common divisor of 68 and 51 is 17. A: the remainder of a natural number divided by 326, 258 and 207 is the same. The natural number is 17