2156
ARITHMETIC
To prove an example in division, multiply the quotient by the divisor, and add the remainder, if there is one.
EXERCISE 7. Divide and prove the following:
1. 6)568
2. 8^9655
3. 9)15432
4. 7)98746
5. 10)76482
6. . 8) 200416
7. 7)4732
8. 4^2116
9. 5)24806
10. 6)85429
11. 9)46681
12. 6)15116
13. 5)5263
14. 3J2004
15. 7)19019
16. 8)34702
17. 5)84763
18. 7)34816
19. 9)6248
20. 6)5084
21. 4)49048
22. 3)15161
23. 2)14631
24. 4)648161
LONG DIVISION.
Divide 8401 by 31.
The process of long division is the same as that of short
271 division, except that in long division we express the sueees-
31)8401 sive partial dividends.
62 We find the quotient figure by comparing the first one, or
the first two dividend figures with the first divisor figure.
220 In the above example:
217 3 is contained in 8, 2 times.
3 is contained in 22, 7 times.
31 3 is contained in 3, once.
31 Thus the quotient is 271.
— Always place the quotient figure directly over the dividend
0 figure that produced it.
Divide 8401 by 82.
Examples where we have zero in the quotient often give trouble.
102J-I
82)8401 82
Practice, however, does away with the difficulty. We find in the above example that 82 from 84 leaves a remainder of 2: bring down the 0. We see that 82 is not contained in 20; write 0 in the quotient directly over the dividend figure which was last brought down, and bring down the next figure 1. 82 is contained in 201 twice. Thus
the quotient is 102 with a remainder of 37.
37 remainder.
If after bringing down another figure the divisor is still larger than the partial dividend, write another 0 in the quotient, and bring down the next quotient figure.
201 164
