Question:
A)  What is the first interger that when squared ends in three 4's? (ex. x^2 = ...444) 
B)  What are all intergers that when squared ends in three 4's? 
C)  prove that there are no intergers that when squared end in four 4's (ex. x^2 = ...4444) 
Suppose n is an integer such that n^2 ends in 444 then the units digit of n^2 is 4 and thus the units digit of n must be 2 or 8.
Suppose the units digit of n is 2.
Write n=100a+10b+2 where a is an integer and b is a single digit, then n^2=10000 a^2 + 2000 ab + 400 a + 100 b^2 + 40 b + 4 Since the ten's digit of n^2 is 4, b must be 1 or 6. 

b=1: 
n^2=(100a+12)^2
=10000a^2 + 2400a + 144. Notice that the hundreds digit of 144 is odd and the hundreds digit of 10000a^2 + 2400a is even so the hundreds digit of their sum, n^2, is odd and thus cannot be 4. 

b=6: 
n^2=(100a+62)^2
=10000a^2 + 12400a + 3844.
The hundreds digit of 10000a^2 + 12400a + 3844, which must be 4, is the units digit of 4a +8 and thus the units digit of a is 4 or 9. Thus we get 
Suppose the units digit of n is 8.
Write n=100a+10b+2 where a is an integer and b is a single digit, then n^2=10000 a^2 + 2000 ab + 1600 a + 100 b^2 + 160 b + 64 Since the ten's digit of n^2 is 4, b must be 3 or 8. 

b=8: 
n^2=(100a+88)^2
=10000 a^2 + 17600 a + 7744. Again the hundreds digit of 7744 is odd and the hundreds digit of 10000a^2 + 17600a is even so the hundreds digit of their sum, n^2, is odd and thus cannot be 4. 

b=3: 
n^2=(100a+38)^2
=10000 a^2 + 7600 a + 1444.
The hundreds digit of 10000 a^2 + 7600 a + 1444, which must be 4, is the units digit of 6a +4 and thus the units digit of a is 0 or 5. Thus we get 
Hence the intergers that when squared ends in three 4's are of the form

Thus the smallest such number is 38.
For part C) n must first satisfy the condition above, thus n = 1000k+s and
n^2 = 1000000 k^2 + 2000 sk + s^2.
Notice that for each of the four possibilities for s the thousands place of s^2 is odd. The thousands digit of
1000000 k^2 + 2000 sk is even and thus the thousands digit of n^2 is odd and cannot be 4.
Cheers,
Penny
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