Determine all positive integers with the property that they are one more than the sum of the squares of their digits in base 10
No single digit numbers can fit these criteria because that would mean, for a one-digit number n,
n = n^{2} + 1
n = (1 + √(-3))/2
which is not a real number, let alone a positive integer!
We now test all two-digit numbers written 10x+y, where x and y are non negative integers (and are less than 10)
.
All numbers that work will be solutions for
x^{2} + y^{2} + 1 = 10x + y
or x^{2} 10x + y^{2} y + 1 = 0
Solving for x, we get
x = (10 + √(100 4(y^{2} y + 1)))/2
= 5 + √(25 y^{2} + y 1)
= 5 + √(24 y(y 1))
= 5 + D where D = √(24 y(y 1))
Trying y = 0, 1, 2, 3, … 9, we find that if y > 5, (24 y(y 1)) becomes negative and there are no real solutions for x. Thus y<6.
.
Trying 0, 1, 2, 3, 4, and 5, we find that the only integer solution is y = 5, and we get D = 2
This gives x = 5 + 2,
or x = 3 or 7.
35 and 75 are the only two-digit answers.
We will now prove that there are no other solutions.
Let N be the number and S be the sum of the squares of its digits plus one.
When N is 999, the largest possible three digit number, S = 244. This means that we need only test up to 244, because for any number above that S will be less than N.
Now since the first digit is either 1 or 2, S is less than (4 + 81 + 81 + 1), or 167. Thus the first digit can only be 1.
Now let the 3-digit number be 100 + 10x + y.
Then,
1 + x^{2} + y^{2} + 1 = 100 + 10x + y
or x^{2} 10x + y^{2} y 98 = 0
Again, solving for x gives
x = (10 + √(100 4(y^{2} y 98)))/2
= 5 + √(25 y^{2} + y + 98)
= 5 + √(123 y(y 1)
= 5 + D where D^{2} = 123 y(y 1)
Now the smallest possible value of D^{2} is when y = 9, and in this case,
D > 7.
But if D > 7, there are no valid values for x , and hence there are no solutions.
(Note: x is non negative and less than 10)
Now, if we increase the number of digits by 1, we add at most 81 to S, but we add at least 1000 to N, and thus we can never have S and N equal. Thus there are no more solutions.
35 and 75 are the only solutions.
Prove that if n is an odd positive integer, then N = 2269^{n} +1779^{n} +
1730^{n} 1776^{n} is an integer multiple of 2001.
In modulo 3:
2269 ≡ 1
1779 ≡ 0
1730 ≡ 2
1776 ≡ 0
So,
N ≡ 1^{n} + 0^{n} + 2^{n} 0^{n}
≡ 1^{n} + (-1)^{n}
Now, if n is an odd positive integer, then a negative number raised to n will be negative, or (-1)^{n} = -1^{n}.
N ≡ 1^{n} 1^{n} ≡ 0
In modulo 23, similarly
2269 ≡ 15 ≡ (-8)
1779 ≡ 8
1730 ≡ 5
1776 ≡ 5
So,
N ≡ 15^{n} + 8^{n} + 5^{n} 5^{n}
≡ (-8)^{n} + 8^{n} + 5^{n} 5^{n}
≡ 8^{n} 8^{n} + 5^{n} 5^{n}
N ≡ 0 (mod 23)
Again, in modulo 29:
2269 ≡ 7
1779 ≡ 10
1730 ≡ 19 ≡ (-10)
1776 ≡ 7
So,
N ≡ 7^{n} + 10^{n} + 19^{n} 7^{n}
≡ 7^{n} + 10^{n} + (-10)^{n} 7^{n}
≡ 7^{n} 7^{n } + 10^{n} 10^{n}
N ≡ 0 (mod 29)
So, N is divisible by 3, 21, and 29, thus N is divisible by 2001.
Hexagon RSTUVW is constructed by starting with a right triangle of legs measuring p and q, constructing squares outwardly on the sides of this triangle, and then connecting the outer vertices of the squares, as shown in the figure below.
Given that p and q are integers with p > q, and that the area of RSTUVW is 1922, determine p and q
We label the figure as shown above. We will determine the area of RSTUVW in terms of p and q.
Triangles ABC and ARW are congruent because corresponding legs are congruent, and both triangles are right. Thus, they have the same areas.
To find the area of ΔUVC, we will draw a line parallel to VC through U. We will draw a perpendicular to this line through C, and call the point of intersection X, forming the right triangle CXV with the same base and height (and, therefore, the same area) as ΔVCU.
ABC measures 90 mACB because CAB is a right angle. XCU also measures 90 mACB because UCB measures 90 and C lies on AX. Thus, XCU is congruent to ABC. Also, UC is congruent to CB because they are sides of the same square. So, by HA, ΔABC is congruent to ΔXCU, and thus the base height of ΔVCX and ΔVCU are equal respectively, and thus they have the same area as ΔABC.
Using the same procedure, we can find the area of ΔTBS by drawing a parallel to BS through T and a perpendicular to this line through B. We will call the intersection of these two lines Y. Using the same method as above, we find that TBY is congruent to BCA, and using HA, we can say that ΔTBS has the same area as ΔABC.
The area of ΔABC is (pq)/2 and since there are four triangles with this area, the area of all the triangular regions of RSTUVW is 2pq.
The area of square ABSR is p^{2}, the area of WRCV is q^{2}, and the area of square CBTU is p^{2} + q^{2} because ΔABC is a right triangle.
The area of RSTUVW in terms of p and q is
2pq + 2(p^{2}+ q^{2})
2pq + 2(p^{2} + q^{2}) = 1922
or p^{2} + pq + q^{2} = 961
As we take p > q,
961 = p^{2} + pq + q^{2} > q^{2} + q(q) + q^{2} = 3q^{2},
or q < √(961/3) < 17
Solving for p, we get
p^{2} + pq + q^{2} 961 = 0
p = (-q + √(q^{2} 4(q^{2} 961))/2
Since (p + q) is positive, we need only consider the positive sign on the radical.
p = (-q + √(3844 3q^{2}))/2
or p = (-q + √(D))/2, where D = √(3844 3q^{2})
Now, since 17 is a small number, we can use brute force, and find which values of integer q give an integer value for D.
q |
D^{2} |
Comments |
1 |
3841 |
Not a perfect square |
2 |
3832 |
Not a perfect square |
3 |
3817 |
Not a perfect square |
4 |
3796 |
Not a perfect square |
5 |
3769 |
Not a perfect square |
6 |
3736 |
Not a perfect square |
7 |
3697 |
Not a perfect square |
8 |
3652 |
Not a perfect square |
9 |
3601 |
Not a perfect square |
10 |
3544 |
Not a perfect square |
11 |
3481 |
= 59^{2} |
12 |
3412 |
Not a perfect square |
13 |
3337 |
Not a perfect square |
14 |
3256 |
Not a perfect square |
15 |
3169 |
Not a perfect square |
16 |
3076 |
Not a perfect square |
17 |
2977 |
Not a perfect square |
Thus, the only integer solution for q is 11, and this gives
p = (-11 + 59)/2
p = 24
p = 24, q = 11