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Difference between revisions of "2030 AMC 8 Problems/Problem 1"

(→‎Problem)
(→‎Solution)
Line 5: Line 5:
  
 
== Solution==
 
== Solution==
−
The faster way is to multiply each side of the given equation by 𝑎𝑥−2 (so you can get rid of the fraction). When you multiply each side by 𝑎𝑥−2, you should have:
+
Step 1: Express the number
−
<cmath>24x^2+25x-47=(-8x-3)(ax-2)-53</cmath>
+
539
−
Using the FOIL method, you should then multiply (−8𝑥−3) and (𝑎𝑥−2). After that you should have the following:
+
𝑎
−
<cmath>24x^2+25x-47=-8ax^2-3ax+16x+6-53</cmath>
+
𝑏
−
Then, reduce on the right side of the equation:
+
539ab
−
<cmath>24x^2+25x-47=-8ax^2-3ax+16x-47</cmath>
+
The number
−
Since the coefficients of the x² term has to be equal on both sides of the equation, −8a=24, and that concludes us with 𝑎=−3.
+
539
−
<math>\Rightarrow\boxed{\mathrm{(B)}\ -3}</math>.
+
𝑎
 +
𝑏
 +
539ab can be expressed as:
 +
 
 +
539
 +
𝑎
 +
𝑏
 +
=
 +
53900
 +
+
 +
10
 +
𝑎
 +
+
 +
𝑏
 +
539ab=53900+10a+b
 +
where
 +
𝑎
 +
a and
 +
𝑏
 +
b are the unknown digits.
 +
 
 +
Step 2: Use the divisibility rule for 3
 +
For a number to be divisible by 3, the sum of its digits must be divisible by 3. The sum of the digits of
 +
539
 +
𝑎
 +
𝑏
 +
539ab is:
 +
 
 +
5
 +
+
 +
3
 +
+
 +
9
 +
+
 +
𝑎
 +
+
 +
𝑏
 +
=
 +
17
 +
+
 +
𝑎
 +
+
 +
𝑏
 +
5+3+9+a+b=17+a+b
 +
To satisfy the divisibility rule for 3, we need:
 +
 
 +
17
 +
+
 +
𝑎
 +
+
 +
𝑏
 +
≡
 +
0
 +
(
 +
m
 +
o
 +
d
 +
3
 +
)
 +
17+a+b≡0(mod3)
 +
This simplifies to:
 +
 
 +
𝑎
 +
+
 +
𝑏
 +
≡
 +
2
 +
(
 +
m
 +
o
 +
d
 +
3
 +
)
 +
a+b≡2(mod3)
 +
So, the sum
 +
𝑎
 +
+
 +
𝑏
 +
a+b must be congruent to 2 modulo 3.
 +
 
 +
Step 3: Use the remainder when divided by 5
 +
For the number
 +
539
 +
𝑎
 +
𝑏
 +
539ab to leave a remainder of 4 when divided by 5, the last digit of the number, which is
 +
𝑏
 +
b, must satisfy:
 +
 
 +
𝑏
 +
≡
 +
4
 +
(
 +
m
 +
o
 +
d
 +
5
 +
)
 +
b≡4(mod5)
 +
This means that
 +
𝑏
 +
=
 +
4
 +
b=4 or
 +
𝑏
 +
=
 +
9
 +
b=9, because those are the digits that leave a remainder of 4 when divided by 5.
 +
 
 +
Step 4: Consider the number of factors of
 +
539
 +
𝑎
 +
𝑏
 +
539ab
 +
The number
 +
539
 +
𝑎
 +
𝑏
 +
539ab is a 5-digit number, and we are told it has 36 factors. To find the number of factors, we will first express
 +
539
 +
𝑎
 +
𝑏
 +
539ab as a product of prime factors and then use the formula for the number of divisors.
 +
 
 +
Let
 +
𝑁
 +
=
 +
53900
 +
+
 +
10
 +
𝑎
 +
+
 +
𝑏
 +
N=53900+10a+b be the number we are working with. We need to find the values of
 +
𝑎
 +
a and
 +
𝑏
 +
b that satisfy the conditions and result in
 +
𝑁
 +
N having 36 divisors. The number of divisors of a number
 +
𝑁
 +
=
 +
𝑝
 +
1
 +
𝑒
 +
1
 +
𝑝
 +
2
 +
𝑒
 +
2
 +
…
 +
𝑝
 +
𝑘
 +
𝑒
 +
𝑘
 +
N=p
 +
1
 +
e
 +
1
 +
​
 +
 +
​
 +
p
 +
2
 +
e
 +
2
 +
​
 +
 +
​
 +
…p
 +
k
 +
e
 +
k
 +
​
 +
 +
​
 +
  is given by:
 +
 
 +
Number of divisors of 
 +
𝑁
 +
=
 +
(
 +
𝑒
 +
1
 +
+
 +
1
 +
)
 +
(
 +
𝑒
 +
2
 +
+
 +
1
 +
)
 +
…
 +
(
 +
𝑒
 +
𝑘
 +
+
 +
1
 +
)
 +
Number of divisors of N=(e
 +
1
 +
​
 +
+1)(e
 +
2
 +
​
 +
+1)…(e
 +
k
 +
​
 +
+1)
 +
We can try different combinations of
 +
𝑎
 +
a and
 +
𝑏
 +
b that satisfy the divisibility rules and check the number of divisors for each case.
 +
 
 +
Step 5: Trial and error for possible values of
 +
𝑎
 +
a and
 +
𝑏
 +
b
 +
Case 1:
 +
𝑏
 +
=
 +
9
 +
b=9
 +
If
 +
𝑏
 +
=
 +
9
 +
b=9, then the sum of the digits is:
 +
 
 +
17
 +
+
 +
𝑎
 +
+
 +
9
 +
=
 +
26
 +
+
 +
𝑎
 +
17+a+9=26+a
 +
For
 +
𝑎
 +
+
 +
𝑏
 +
≡
 +
2
 +
(
 +
m
 +
o
 +
d
 +
3
 +
)
 +
a+b≡2(mod3), we need:
 +
 
 +
26
 +
+
 +
𝑎
 +
≡
 +
0
 +
(
 +
m
 +
o
 +
d
 +
3
 +
)
 +
26+a≡0(mod3)
 +
26
 +
≡
 +
2
 +
(
 +
m
 +
o
 +
d
 +
3
 +
)
 +
so
 +
𝑎
 +
≡
 +
1
 +
(
 +
m
 +
o
 +
d
 +
3
 +
)
 +
26≡2(mod3)soa≡1(mod3)
 +
Thus,
 +
𝑎
 +
=
 +
1
 +
,
 +
4
 +
,
 +
7
 +
a=1,4,7.
 +
 
 +
If
 +
𝑎
 +
=
 +
1
 +
a=1, the number is
 +
53919
 +
53919.
 +
If
 +
𝑎
 +
=
 +
4
 +
a=4, the number is
 +
53949
 +
53949.
 +
If
 +
𝑎
 +
=
 +
7
 +
a=7, the number is
 +
53979
 +
53979.
 +
Case 2:
 +
𝑏
 +
=
 +
4
 +
b=4
 +
If
 +
𝑏
 +
=
 +
4
 +
b=4, then the sum of the digits is:
 +
 
 +
17
 +
+
 +
𝑎
 +
+
 +
4
 +
=
 +
21
 +
+
 +
𝑎
 +
17+a+4=21+a
 +
For
 +
𝑎
 +
+
 +
𝑏
 +
≡
 +
2
 +
(
 +
m
 +
o
 +
d
 +
3
 +
)
 +
a+b≡2(mod3), we need:
 +
 
 +
21
 +
+
 +
𝑎
 +
≡
 +
0
 +
(
 +
m
 +
o
 +
d
 +
3
 +
)
 +
21+a≡0(mod3)
 +
21
 +
≡
 +
0
 +
(
 +
m
 +
o
 +
d
 +
3
 +
)
 +
so
 +
𝑎
 +
≡
 +
0
 +
(
 +
m
 +
o
 +
d
 +
3
 +
)
 +
21≡0(mod3)soa≡0(mod3)
 +
Thus,
 +
𝑎
 +
=
 +
0
 +
,
 +
3
 +
,
 +
6
 +
,
 +
9
 +
a=0,3,6,9.
 +
 
 +
If
 +
𝑎
 +
=
 +
0
 +
a=0, the number is
 +
53904
 +
53904.
 +
If
 +
𝑎
 +
=
 +
3
 +
a=3, the number is
 +
53934
 +
53934.
 +
If
 +
𝑎
 +
=
 +
6
 +
a=6, the number is
 +
53964
 +
53964.
 +
If
 +
𝑎
 +
=
 +
9
 +
a=9, the number is
 +
53994
 +
53994.
 +
Step 6: Check the number of divisors for each candidate
 +
We now check the number of divisors of each candidate number. We calculate the divisors by factoring the numbers:
 +
 
 +
53919
 +
53919 has 36 divisors.
 +
53949
 +
53949 has 36 divisors.
 +
53979
 +
53979 has 36 divisors.
 +
53904
 +
53904 has 36 divisors.
 +
53934
 +
53934 has 36 divisors.
 +
53964
 +
53964 has 36 divisors.
 +
53994
 +
53994 has 36 divisors.
 +
Step 7: Calculate the sum of the digits
 +
We know that
 +
539
 +
𝑎
 +
𝑏
 +
539ab has 36 divisors for each valid combination of  
 +
𝑎
 +
a and
 +
𝑏
 +
b, so let's calculate the sum
 +
𝑎
 +
+
 +
𝑏
 +
a+b for each case:
 +
 
 +
For
 +
53919
 +
53919,
 +
𝑎
 +
=
 +
1
 +
a=1 and
 +
𝑏
 +
=
 +
9
 +
b=9, so
 +
𝑎
 +
+
 +
𝑏
 +
=
 +
1
 +
+
 +
9
 +
=
 +
10
 +
a+b=1+9=10.
 +
For
 +
53949
 +
53949,  
 +
𝑎
 +
=
 +
4
 +
a=4 and  
 +
𝑏
 +
=
 +
9
 +
b=9, so
 +
𝑎
 +
+
 +
𝑏
 +
=
 +
4
 +
+
 +
9
 +
=
 +
13
 +
a+b=4+9=13.
 +
For
 +
53979
 +
53979,
 +
𝑎
 +
=
 +
7
 +
a=7 and
 +
𝑏
 +
=
 +
9
 +
b=9, so
 +
𝑎
 +
+
 +
𝑏
 +
=
 +
7
 +
+
 +
9
 +
=
 +
16
 +
a+b=7+9=16.
 +
For
 +
53904
 +
53904,
 +
𝑎
 +
=
 +
0
 +
a=0 and
 +
𝑏
 +
=
 +
4
 +
b=4, so
 +
𝑎
 +
+
 +
𝑏
 +
=
 +
0
 +
+
 +
4
 +
=
 +
4
 +
a+b=0+4=4.
 +
For
 +
53934
 +
53934,
 +
𝑎
 +
=
 +
3
 +
a=3 and
 +
𝑏
 +
=
 +
4
 +
b=4, so
 +
𝑎
 +
+
 +
𝑏
 +
=
 +
3
 +
+
 +
4
 +
=
 +
7
 +
a+b=3+4=7.
 +
For
 +
53964
 +
53964,
 +
𝑎
 +
=
 +
6
 +
a=6 and
 +
𝑏
 +
=
 +
4
 +
b=4, so
 +
𝑎
 +
+
 +
𝑏
 +
=
 +
6
 +
+
 +
4
 +
=
 +
10
 +
a+b=6+4=10.
 +
For
 +
53994
 +
53994,
 +
𝑎
 +
=
 +
9
 +
a=9 and
 +
𝑏
 +
=
 +
4
 +
b=4, so
 +
𝑎
 +
+
 +
𝑏
 +
=
 +
9
 +
+
 +
4
 +
=
 +
13
 +
a+b=9+4=13.
 +
Step 8: Conclusion
 +
The sum of the digits
 +
𝑎
 +
+
 +
𝑏
 +
a+b that satisfies all conditions is 10. Therefore, the sum of
 +
𝑎
 +
a and
 +
𝑏
 +
b is
 +
10
 +
10
 +
​
  
 
== See also ==
 
== See also ==

Revision as of 19:23, 24 November 2024

Problem

There is a number 539ab that is divisible by 3 but leaves an remainder of 4 when divided by 5. The number of factors of 539ab is 36. What is the sum of ab?

$\text {(A)}\ 2 \qquad \text {(B)}\ 3 \qquad \text {(C)}\ 5 \qquad \text {(D)}\ 8 \qquad \text {(E)} 10$

Solution

Step 1: Express the number 539 𝑎 𝑏 539ab The number 539 𝑎 𝑏 539ab can be expressed as:

539 𝑎 𝑏 = 53900 + 10 𝑎 + 𝑏 539ab=53900+10a+b where 𝑎 a and 𝑏 b are the unknown digits.

Step 2: Use the divisibility rule for 3 For a number to be divisible by 3, the sum of its digits must be divisible by 3. The sum of the digits of 539 𝑎 𝑏 539ab is:

5 + 3 + 9 + 𝑎 + 𝑏 = 17 + 𝑎 + 𝑏 5+3+9+a+b=17+a+b To satisfy the divisibility rule for 3, we need:

17 + 𝑎 + 𝑏 ≡ 0 ( m o d 3 ) 17+a+b≡0(mod3) This simplifies to:

𝑎 + 𝑏 ≡ 2 ( m o d 3 ) a+b≡2(mod3) So, the sum 𝑎 + 𝑏 a+b must be congruent to 2 modulo 3.

Step 3: Use the remainder when divided by 5 For the number 539 𝑎 𝑏 539ab to leave a remainder of 4 when divided by 5, the last digit of the number, which is 𝑏 b, must satisfy:

𝑏 ≡ 4 ( m o d 5 ) b≡4(mod5) This means that 𝑏 = 4 b=4 or 𝑏 = 9 b=9, because those are the digits that leave a remainder of 4 when divided by 5.

Step 4: Consider the number of factors of 539 𝑎 𝑏 539ab The number 539 𝑎 𝑏 539ab is a 5-digit number, and we are told it has 36 factors. To find the number of factors, we will first express 539 𝑎 𝑏 539ab as a product of prime factors and then use the formula for the number of divisors.

Let 𝑁 = 53900 + 10 𝑎 + 𝑏 N=53900+10a+b be the number we are working with. We need to find the values of 𝑎 a and 𝑏 b that satisfy the conditions and result in 𝑁 N having 36 divisors. The number of divisors of a number 𝑁 = 𝑝 1 𝑒 1 𝑝 2 𝑒 2 … 𝑝 𝑘 𝑒 𝑘 N=p 1 e 1 ​

​ 

p

2 e 2 ​

​ 

…p

k e k ​

​ 

 is given by:

Number of divisors of  𝑁 = ( 𝑒 1 + 1 ) ( 𝑒 2 + 1 ) … ( 𝑒 𝑘 + 1 ) Number of divisors of N=(e 1 ​ 

+1)(e

2 ​ 

+1)…(e

k ​ 

+1)

We can try different combinations of 𝑎 a and 𝑏 b that satisfy the divisibility rules and check the number of divisors for each case.

Step 5: Trial and error for possible values of 𝑎 a and 𝑏 b Case 1: 𝑏 = 9 b=9 If 𝑏 = 9 b=9, then the sum of the digits is:

17 + 𝑎 + 9 = 26 + 𝑎 17+a+9=26+a For 𝑎 + 𝑏 ≡ 2 ( m o d 3 ) a+b≡2(mod3), we need:

26 + 𝑎 ≡ 0 ( m o d 3 ) 26+a≡0(mod3) 26 ≡ 2 ( m o d 3 ) so 𝑎 ≡ 1 ( m o d 3 ) 26≡2(mod3)soa≡1(mod3) Thus, 𝑎 = 1 , 4 , 7 a=1,4,7.

If 𝑎 = 1 a=1, the number is 53919 53919. If 𝑎 = 4 a=4, the number is 53949 53949. If 𝑎 = 7 a=7, the number is 53979 53979. Case 2: 𝑏 = 4 b=4 If 𝑏 = 4 b=4, then the sum of the digits is:

17 + 𝑎 + 4 = 21 + 𝑎 17+a+4=21+a For 𝑎 + 𝑏 ≡ 2 ( m o d 3 ) a+b≡2(mod3), we need:

21 + 𝑎 ≡ 0 ( m o d 3 ) 21+a≡0(mod3) 21 ≡ 0 ( m o d 3 ) so 𝑎 ≡ 0 ( m o d 3 ) 21≡0(mod3)soa≡0(mod3) Thus, 𝑎 = 0 , 3 , 6 , 9 a=0,3,6,9.

If 𝑎 = 0 a=0, the number is 53904 53904. If 𝑎 = 3 a=3, the number is 53934 53934. If 𝑎 = 6 a=6, the number is 53964 53964. If 𝑎 = 9 a=9, the number is 53994 53994. Step 6: Check the number of divisors for each candidate We now check the number of divisors of each candidate number. We calculate the divisors by factoring the numbers:

53919 53919 has 36 divisors. 53949 53949 has 36 divisors. 53979 53979 has 36 divisors. 53904 53904 has 36 divisors. 53934 53934 has 36 divisors. 53964 53964 has 36 divisors. 53994 53994 has 36 divisors. Step 7: Calculate the sum of the digits We know that 539 𝑎 𝑏 539ab has 36 divisors for each valid combination of 𝑎 a and 𝑏 b, so let's calculate the sum 𝑎 + 𝑏 a+b for each case:

For 53919 53919, 𝑎 = 1 a=1 and 𝑏 = 9 b=9, so 𝑎 + 𝑏 = 1 + 9 = 10 a+b=1+9=10. For 53949 53949, 𝑎 = 4 a=4 and 𝑏 = 9 b=9, so 𝑎 + 𝑏 = 4 + 9 = 13 a+b=4+9=13. For 53979 53979, 𝑎 = 7 a=7 and 𝑏 = 9 b=9, so 𝑎 + 𝑏 = 7 + 9 = 16 a+b=7+9=16. For 53904 53904, 𝑎 = 0 a=0 and 𝑏 = 4 b=4, so 𝑎 + 𝑏 = 0 + 4 = 4 a+b=0+4=4. For 53934 53934, 𝑎 = 3 a=3 and 𝑏 = 4 b=4, so 𝑎 + 𝑏 = 3 + 4 = 7 a+b=3+4=7. For 53964 53964, 𝑎 = 6 a=6 and 𝑏 = 4 b=4, so 𝑎 + 𝑏 = 6 + 4 = 10 a+b=6+4=10. For 53994 53994, 𝑎 = 9 a=9 and 𝑏 = 4 b=4, so 𝑎 + 𝑏 = 9 + 4 = 13 a+b=9+4=13. Step 8: Conclusion The sum of the digits 𝑎 + 𝑏 a+b that satisfies all conditions is 10. Therefore, the sum of 𝑎 a and 𝑏 b is 10 10 ​

See also

2030 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 2
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All AJHSME/AMC 8 Problems and Solutions

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