x^2 = ...444

Date: Sat, 21 Feb 1998 19:16:13 -0600 (CST)
Name: James
Who is asking: Student
Level: Secondary

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)

Hi James

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
n=1000k + 462 or n=1000k + 962 where k is a positive integer.

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
n=1000k + 38 or n=1000k + 538 where k is a positive integer.

Hence the intergers that when squared ends in three 4's are of the form

1000k+s

where k is a positive integer and s is 38, 462, 538 or 962.

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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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)

	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 n=1000k + 462 or n=1000k + 962 where k is a positive integer.

	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 n=1000k + 38 or n=1000k + 538 where k is a positive integer.