[Prime]

While surfing the Net, I came upon a sample admission test in math for the MEPhI (Moscow Engineering Physics Institute). As I was looking at those 5 problems, a profound sense of futility slowly rose from the pit of my stomach. But upon a closer examination I realized that the problem #2 I could solve and quickly enough.

Here is my English translation thereof. Enjoy.

2. If you multiply the first digit of a certain two-digit number by 4, the second digit by 6 and add the results, then the resulting sum is going to be 13 greater than the sum of the squares of those two digits.

Find such two-digit number.

That is the easiest (or second easiest) of the 5 problems. The other four I do not want to show here, as I find them unsuitable.

[Guest]

Let the number is ab, after making the statement:

a^2 - 14a + b^2 - 7b = -13

a(a-14) + b(b-7) = -13

There are two major elements at the left side. If I equalize one element to -13 and the other to zero, it will work.

a(a-14)=-13 --> a=1

b(b-7)=0 ----> b=7

[Guest]

While surfing the Net, I came upon a sample admission test in math for the MEPhI (Moscow Engineering Physics Institute). As I was looking at those 5 problems, a profound sense of futility slowly rose from the pit of my stomach. But upon a closer examination I realized that the problem #2 I could solve and quickly enough.

Here is my English translation thereof. Enjoy.

2. If you multiply the first digit of a certain two-digit number by 4, the second digit by 6 and add the results, then the resulting sum is going to be 13 greater than the sum of the squares of those two digits.

Find such two-digit number.

That is the easiest (or second easiest) of the 5 problems. The other four I do not want to show here, as I find them unsuitable.

32

By trial and error... dunno if there's an easy way to get there mathematically

[bonanova]

23

[Guest]

Let the number is ab, after making the statement:

a^2 - 14a + b^2 - 7b = -13

a(a-14) + b(b-7) = -13

There are two major elements at the left side. If I equalize one element to -13 and the other to zero, it will work.

a(a-14)=-13 --> a=1

b(b-7)=0 ----> b=7

a^2 -4a + b^2 -6b = -13

[Prime]

23

You have the correct answer. But this wouldn't pass. (Or you'd get a small partial credit.) They want to see your work. This problem must be solved analytically, not just by trial and error.

[Guest]

[23

4x+6y=x^2+y^2+13

so,(x-2)^2+(y-3)^2=0

so,number is 23

Edited November 26, 2008 by bonanova
spoler added to protect the puzzle solvers

[bonanova]

1. Because the sum of products with even numbers differs from sum of squares by an odd number, one digit is even, the other is odd.
   
2. Because the digits multiplied by 4 and 6 are larger then their squares, the digits themselves are smaller than 4 and 6.
   
3. Since 13 is the square of a number between 3 and 4, the digits are smaller than 4 and 6 by about 3 or 4.
   

Pretty small search space.

I tried 12, 21, 23 and stopped; because 23 worked.

What I did not do was show - as subhobrata did in the prior post - that the answer is unique.

Since the OP asks only to find such [a] number, he should be awarded extra credit.

[Prime]

[23

4x+6y=x^2+y^2+13

so,(x-2)^2+(y-3)^2=0

so,number is 23

That's the solution. Also, partial credit to Armcie for same correct equation.

Let me show a bit more detailed work.

4x + 6y = x² + y² + 13

x² - 4x + 4 + y² - 6x + 9 =0

(x-2)² + (y-3)² = 0

And we are looking for the real number, integer solutions. So x=2; y=3. The number is 23.

Because the sum of products with even numbers differs from sum of squares by an odd number, one digit is even, the other is odd.

Because the digits multiplied by 4 and 6 are larger then their squares, the digits themselves are smaller than 4 and 6.

Since 13 is the square of a number between 3 and 4, the digits are smaller than 4 and 6 by about 3 or 4.

I tried 12, 21, 23 and stopped; because 23 worked.

That is part analytical, part trial and error. Also, the even digit could be zero.

[bonanova]

Also, the even digit could be zero.

Really? Then what would the odd digit be? Or are you advising that zero is an even number?

Not to belabor the point, but the OP asks to find a number that fit the criteria.

If someone found 23 lying on the sidewalk and gave it to you, they satisfied the OP.

[Prime]

Really? Then what would the odd digit be? Or are you advising that zero is an even number?

Not to belabor the point, but the OP asks to find a number that fit the criteria.

If someone found 23 lying on the sidewalk and gave it to you, they satisfied the OP.

Within the Brainden framework, your solution (by trial and error) is absolutely correct. And trial and error within some defined boundary conditions is a valid analytical method. On that test, I am sure, they wouldn’t give you a full credit, as they looked for an algebraic solution there. Your method does not prove that 23 is the only solution, but that’s not a flaw, since my statement of the problem (my translation rather) in no way implied that requirement.

I mentioned zero, because I didn’t see in your trial and error solution test cases for it. Like number 30, for example, which falls within the parameters that you specified.

I posted that problem as a curiosity item. To show what kind of admission tests they give in that University in Russia. Keep in mind, there is a time limit on the admission test. Even so, that problem was meant as a gift. Others looked much worse.

[Prime]

Here is the Problem 1 from that test. This one is not all that difficult, also.

1. Solve the inequality:

log_(2x+7)(x²+4x+4) < 2log_(x²)|x|

[Guest]

I'm not good in solving problems here, but

-3<x<1

[Prime]

I'm not good in solving problems here, but

-3<x<1

It is in the general neighborhood of the solution. However, they don't give multiple choice test in that university. You must show how you arrive to the answer.

To solve, you need to simplify the expression somewhat. Also, you must mind the values of x, where log is undefined.

[Guest]

It is in the general neighborhood of the solution. However, they don't give multiple choice test in that university. You must show how you arrive to the answer.

To solve, you need to simplify the expression somewhat. Also, you must mind the values of x, where log is undefined.

log_(2x+7)(x²+4x+4) <2log_(x²)|x|

2log_(x²) |x| =log_(x²) |x|² = 1

log_(2x+7)(x²+4x+4) < 1 -----> (2x+7)¹ < x²+4x+4

2x+7<x²+4x+4

x²+2x-3>0

(x-1)(x+3)>0

if you consider this function as a graphic, the curve will intersect with zero line at x=1 and x=-3 , and between these values, all x values will proof the statement. Then solution is: -3<x<1. Sure I must add that "except x=0", where log is undefined.

As you guess, I'm far from calculus, and not hopeful that I'm correct. If this solution is correct, I will be astonished, because any high school student would manage it.

[Guest]

log_(2x+7)(x²+4x+4) <2log_(x²)|x|

2log_(x²) |x| =log_(x²) |x|² = 1

log_(2x+7)(x²+4x+4) < 1 -----> (2x+7)¹ < x²+4x+4

2x+7<x²+4x+4

x²+2x-3>0

(x-1)(x+3)>0

if you consider this function as a graphic, the curve will intersect with zero line at x=1 and x=-3 , and between these values, all x values will proof the statement. Then solution is: -3<x<1. Sure I must add that "except x=0", where log is undefined.

As you guess, I'm far from calculus, and not hopeful that I'm correct. If this solution is correct, I will be astonished, because any high school student would manage it.

I think you missed x= -2 where log_(2x+7)(x²+4x+4) is undefined.

And I don't know about 'any high school student' being able to manage that... from memory logs were a closed book to many of my colleagues ten years ago. I think the third line would be a loggical step too far for many.

[Guest]

I think you missed x= -2 where log_(2x+7)(x²+4x+4) is undefined.

And I don't know about 'any high school student' being able to manage that... from memory logs were a closed book to many of my colleagues ten years ago. I think the third line would be a loggical step too far for many.

Yes, I acted hasty and didn't check all the undefined conditions, sorry.

I love math, but I didn't get any further education on math after high school. Although I did it with small mistakes, if I had tried throughly, I would manage it by my knowledge of math remnants from high school.

[Guest]

log_(2x+7)(x²+4x+4) <2log_(x²)|x|

2log_(x²) |x| =log_(x²) |x|² = 1

log_(2x+7)(x²+4x+4) < 1 -----> (2x+7)¹ < x²+4x+4

2x+7<x²+4x+4

x²+2x-3>0

(x-1)(x+3)>0

if you consider this function as a graphic, the curve will intersect with zero line at x=1 and x=-3 , and between these values, all x values will proof the statement. Then solution is: -3<x<1. Sure I must add that "except x=0", where log is undefined.

As you guess, I'm far from calculus, and not hopeful that I'm correct. If this solution is correct, I will be astonished, because any high school student would manage it.

You got something backwards there:

If log_a(b) < 1, then 0<b<a, as opposed to a<b.

[Prime]

We are moving toward the solution, but no solution yet. I am going to make couple hints here, which should help solving this problem properly.

1. When making substitution log_(x²)|x|² = 1, we assume x² > 0 and x² <> 1 (not equal).

2. If log_ab < 1, then

for 0 < a < 1 ---> b > a;

for a > 1 ---> 0 < b < a.

The two problems shown in this post are the easier problems on that test. I don’t think they consider those as calculus problems. Those two are arithmetic and algebra problems.

Problem 3 on that test is a calculus problem. It involves a basic derivative and integral of a polynomial. But the setup leads to a rather complex and lengthy solution.

Problem 4 is a trigonometry problem involving knowledge of some basic formulas, like cosine of doubled angle and other behavior of trigonometric functions. But the setup, again, leads to a very lengthy and complex solution requiring testing of many different cases. So I don’t think it is worth showing those here. (Too much work solving them).

However, problem 5, a geometry problem, I could show when and if the logarithm is solved.