SOLUTION: I need help with this, please and thank you
Let n=179,668,3de be a base-ten numeral with d and e its last two digits. Give all of the choices of the two-digit numb
Algebra ->
Divisibility and Prime Numbers
-> SOLUTION: I need help with this, please and thank you
Let n=179,668,3de be a base-ten numeral with d and e its last two digits. Give all of the choices of the two-digit numb
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Let n=179,668,3de be a base-ten numeral with d and e its last two digits. Give all of the choices of the two-digit numbers de for which n is divisible by 12. Answer by greenestamps(13206) (Show Source):
The prime factorization of the divisor 12 is (2^2)*(3^1). So a number is divisible by 12 if and only if it is divisible by both 3 and 4.
A number is divisible by 4 if the last 2 digits are divisible by 4; a number is divisible by 3 if the sum of the digits is divisible by 3.
The sum of the digits is 40 + (d+e). So look at all values for the digits de that are divisible by 4 and find the ones for which the sum of all the digits is divisible by 3.
de = 04; 40+0+4 = 44 not divisible by 3
de = 08; 40+0+8 = 48 YES, divisible by 3
de = 12; 40+1+2 = 43 not divisible by 3
...
de = 92; 40+9+2 = 51 YES, divisible by 3
de = 96; 40+9+6 = 55 not divisible by 3
So "08" and "92" are two of the solutions; you can look for others.
