How many 4-digit multiples of 21 and 15 are there and 4-digit multiples of either 21 or 15 are there?
1 answer:
Answer:
943
Step-by-step explanation:
21| 1000
| ----------
47 | 13
21-13=8
1000+8=1008
21|9999
|---------
476|3
9999-3=9996
1008,1029,...,9996
let n be number of terms
a=1008,d=21,l=9996
9996=1008+(n-1)21
9996-1008=(n-1)21
8988=(n-1)21
n-1=8988/21
n-1=428
n=428+1=429
again 15|1000
|---------
66|10
15-10=5
1000+5=1005
15|9999
| ---------
666|9
9999-9=9990
1005,1020,...,9990
a=1005,d=15,l=9990
let n be number of terms.
9990=1005+(n-1)15
9990-1005=(n-1)15
8985=(n-1)15
n-1=8985/15
n-1=599
n=599+1=600
21=3×7
15=3×5
L.C.M.=3×5×7=105
105)1000(9
-945
____
55
105-55=50
1000+50=1050
105)9999(
-945
---------
549
- 525
-------
24
-------
9999-24=9975
1050,1155,...,9975
let n be number of terms.
d=105
l=a+(n-1)d
9975=1050+(n-1)105
9975-1050=(n-1)105
8025=(n-1)105
n-1=8925/105
n-1=85
n=85+1=86
so there are 86 multiples of 21 and 15
(A∪B)=A+B-(A∩B)
multiples of 21 or 15=429+600-86=1029-86=943
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