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katovenus [111]
3 years ago
9

WILL CHOOSE BRAINIEST How many 6-digit numbers, formed using each digit from 1 to 6 exactly once, are divisible by: a.)5 b.)6 c.

)12 d.)25
Mathematics
2 answers:
Natasha_Volkova [10]3 years ago
6 0
D) 25 tell me if I did it righ
andrew11 [14]3 years ago
6 0

Answer:

<u>a) divisible by  5:  120 numbers</u>

<u>b) divisible by  6:    360 numbers</u>

<u>c) divisible by  12:     192 numbers</u>

<u>d) divisible by  25:    24 numbers</u>

Step-by-step explanation:

6-digit  numbers => the numbers have the form abcdef

a.) <u>divisible by  5</u>

f = {5}

a  <u>5 values</u>;   b  <u>4 values</u>, c  <u>3 values</u>,  d  <u>2 values</u> ,   e  <u>1 value</u><em>, </em>f <em> </em><u>1 value</u>

=> 5 * 4 * 3 * 2 * 1 * 1 =<u> 120 numbers</u>

b.)<u> divisible by  6</u>

f = {2; 4; 6}

a  <u>5 values</u>;   b  <u>4 values</u>, c  <u>3 values</u>,  d   <u>2 values</u>,   e <u>1 value</u>, f  <u>3 values</u>

=> 5 * 4 * 3 * 2 * 1 * 3 = <u>360 numbers</u>

c.)<u> divisible by  12</u>

<u>ef</u> =  {12; 16; 24; 32; 36; 52; 56; 64}

a  <u>4 values</u>;   b  <u>3 values</u>, c  <u>2 values</u>,  d   <u>1 value</u>,   <u>ef</u>  <u>8 values</u>

=> 4 * 3 * 2 * 1 * 8 =<u>192 numbers</u>

d.)<u> divisible by  25</u>

<u>ef</u> =  {25}

a  <u>4 values</u>;   b  3<u> values</u>, c  <u>2 values</u>,  d   1 values,   <u>ef</u>  <u>1 value</u>

=> 4 * 3 * 2 *1*1 = <u>24 numbers</u>

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