buoyant wrote:How many two-digit whole numbers yield a remainder of 1 when divided by 10 and also yield a remainder of 1 when divided by 6?
a) None
b) One
c) Two
d) Three
e) Four
Let x = the two-digit integer.
x has a remainder of 1 when divided by 10.
In other words, x is 1 more than a multiple of 10:
x = 10a + 1, where a is an nonnegative integer.
x has a remainder of 1 when divided by 6.
In other words, x is 1 more than a multiple of 6:
x = 6b + 1, where b is a nonnegative integer.
Since x = 10a + 1 and x = 6b + 1, we get;
10a + 1 = 6b + 1
10a = 6b
5a = 3b
a = (3/5)b.
If b=5, then a=3, in which case x = 10a + 1 = 10*3 + 1 = 31.
If b=10, then a=6, in which case x = 10a + 1 = 10*6 + 1 = 61.
If b=15, then a=9, in which case x = 10a + 1 = 10*9 + 1 = 91.
A total of 3 options for x:
31, 61, and 91.
The correct answer is
D.
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