Question

disvibity test 1 to 7 example​

Answer 1

Answer:

1 No special condition. Any integer is divisible by 1. 2 is divisible by 1.

2 The last digit is even (0, 2, 4, 6, or 8).[2][3] 1294: 4 is even.

3 Sum the digits. The result must be divisible by 3.[2][4][5] 405 → 4 + 0 + 5 = 9 and 636 → 6 + 3 + 6 = 15 which both are clearly divisible by 3.

16,499,205,854,376 → 1+6+4+9+9+2+0+5+8+5+4+3+7+6 sums to 69 → 6 + 9 = 15 → 1 + 5 = 6, which is clearly divisible by 3.

Subtract the quantity of the digits 2, 5, and 8 in the number from the quantity of the digits 1, 4, and 7 in the number. The result must be divisible by 3. Using the example above: 16,499,205,854,376 has four of the digits 1, 4 and 7 and four of the digits 2, 5 and 8; ∴ Since 4 − 4 = 0 is a multiple of 3, the number 16,499,205,854,376 is divisible by 3.

4 The last two digits form a number that is divisible by 4.[2][3] 40,832: 32 is divisible by 4.

If the tens digit is even, the ones digit must be 0, 4, or 8.

If the tens digit is odd, the ones digit must be 2 or 6. 40,832: 3 is odd, and the last digit is 2.

Twice the tens digit, plus the ones digit is divisible by 4. 40832: 2 × 3 + 2 = 8, which is divisible by 4.

5 The last digit is 0 or 5.[2][3] 495: the last digit is 5.

6 It is divisible by 2 and by 3.[6] 1458: 1 + 4 + 5 + 8 = 18, so it is divisible by 3 and the last digit is even, hence the number is divisible by 6.

7 Forming an alternating sum of blocks of three from right to left gives a multiple of 7[5][7] 1,369,851: 851 − 369 + 1 = 483 = 7 × 69

Adding 5 times the last digit to the rest gives a multiple of 7. (Works because 49 is divisible by 7.) 483: 48 + (3 × 5) = 63 = 7 × 9.

Subtracting 2 times the last digit from the rest gives a multiple of 7. (Works because 21 is divisible by 7.) 483: 48 − (3 × 2) = 42 = 7 × 6.

Subtracting 9 times the last digit from the rest gives a multiple of 7. 483: 48 − (3 × 9) = 21 = 7 × 3.

Adding 3 times the first digit to the next and then writing the rest gives a multiple of 7. (This works because 10a + b − 7a = 3a + b; the last number has the same remainder as 10a + b.) 483: 4×3 + 8 = 20,

203: 2×3 + 0 = 6, 63: 6×3 + 3 = 21.

Adding the last two digits to twice the rest gives a multiple of 7. (Works because 98 is divisible by 7.) 483,595: 95 + (2 × 4835) = 9765: 65 + (2 × 97) = 259: 59 + (2 × 2) = 63.

Multiply each digit (from right to left) by the digit in the corresponding position in this pattern (from left to right): 1, 3, 2, -1, -3, -2 (repeating for digits beyond the hundred-thousands place). Adding the results gives a multiple of 7. 483,595: (4 × (-2)) + (8 × (-3)) + (3 × (-1)) + (5 × 2) + (9 × 3) + (5 × 1) = 7.

Compute the remainder of each digit pair (from right to left) when divided by 7. Multiply the rightmost remainder by 1, the next to the left by 2 and the next by 4, repeating the pattern for digit pairs beyond the hundred-thousands place. Adding the results gives a multiple of 7. 194,536: 19|45|36 ; (5x4) + (3x2) + (1x1) = 27, so it is not divisible by 7

204,540: 20|45|40 ; (6x4) + (3x2) + (5x1) = 35, so it is divisible by 7

Step-by-step explanation:

Answer 2

Answer:

1. No special condition. Any integer is divisible by 1.
2. The last digit is even (0, 2, 4, 6, or 8)
3. Sum the digits. The result must be divisible by 3  
4. The last two digits form a number that is divisible by 4.
5. The last digit is 0 or 5.
6. It is divisible by 2 and by 3.
7. Adding 5 times the last digit to the rest gives a multiple of 7. (Works because 49 is divisible by 7.)