7/25/2019 Divisiblility rule

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Divisible

by: If: Examples:

2 The last digit is even (0,2,4,6,8)128is

129is not

3 The sum of the digits is divisible by 3

381 (3+8+1=12, and 123 = 4) Yes

21! (2+1+!=10, and 103 =

3 1"3)No

4 The last 2 digits a#e divisible by 4

1312is (124=3)

!019is not

$ The last digit is 0 o# $1!5is

809is not

6 The numbe# is divisible by both 2 and3

114 (it is even, and 1+1+4=6 and

63 = 2) Yes

308 (it is even, but 3+0+8=11 and

113 = 3 2"3) No

!

%f you double the last digit and subt#a&t it f#om

the #est of the numbe# and the ans'e# is

0, o#

divisible by 7

(ote you &an a**ly this #ule to that ans'e#

again if you 'ant)

6!2 (ouble 2 is 4, 6!4=63, and

63!=-) Yes

-0$ (ouble $ is 10, -010=80,

and 80!=11 3"!) No

8 The last th#ee digits a#e divisible by 8

10-816(8168=102) Yes

216302(3028=3! 3"4) No

7/25/2019 Divisiblility rule

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-

The sum of the digits is divisible by -

(ote you &an a**ly this #ule to that ans'e#

again if you 'ant)

162- (1+6+2+-=18, and again,

1+8=-) Yes

2013 (2+0+1+3=6) No

10 The numbe# ends in 0220is

221is not

11

.dd and subt#a&t digits in an alte#nating *atte#n

(add fi#st, subt#a&t se&ond, add thi#d, et&)/ Then

the ans'e# must be

0, o#

divisible by 11

1364 (13+64 = 0) Yes

913(-1+3 = 11) Yes

3!2- (3!+2- = 11) Yes

987(-8+! = 8) No

12 The numbe# is divisible by both 3 and4

648

(By 3?6+4+8=18 and 183=6

es/

By 4?484=12 es) Yes

$24

(By 3?$+2+4=11, 113= 3 2"3o/

ont need to &he& by 4/) No