Difference between revisions of "1998 AHSME Problems/Problem 30"

Latest revision as of 09:37, 11 August 2020

Problem

For each positive integer $n$ , let

$a_n = \frac{(n+9)!}{(n-1)!}$

Let $k$ denote the smallest positive integer for which the rightmost nonzero digit of $a_k$ is odd. The rightmost nonzero digit of $a_k$ is

$\mathrm{(A) \ }1 \qquad \mathrm{(B) \ }3 \qquad \mathrm{(C) \ }5 \qquad \mathrm{(D) \ } 7 \qquad \mathrm{(E) \ } 9$

Solution

We have $a_n = n(n+1)\dots (n+9)$ .

The value $a_n$ can be written as $2^{x_n} 5^{y_n} r_n$ , where $r_n$ is not divisible by 2 and 5. The number of trailing zeroes is $z_n = \min(x_n,y_n)$ . The last non-zero digit is the last digit of $2^{x_n-z_n} 5^{y_n-z_n} r_n$ .

Clearly, the last non-zero digit is even iff $x_n - z_n > 0$ iff $x_n > y_n$ .

Thus we are looking for the smallest $n$ such that the power of $5$ that divides $a_n$ is at least equal to the power of $2$ that divides $a_n$ .

The number $a_n$ is a product of $10$ consecutive integers. Out of these, $5$ are divisible by $2$ . Out of those $5$ , at least $2$ are divisible by $4$ , and out of those $2$ , one is divisible by $8$ . Therefore $x_n\geq 5+2+1=8$ for all $n$ .

On the other hand, exactly $2$ of our ten integers are divisible by $5$ , and at most one of them can be divisible by a higher power of $5$ . As we need $y_n\geq x_n\geq 8$ , one of the integers from $n$ to $n+9$ must be divisible by $5^7 = 78125$ . Therefore $n\geq 78116$ .

We can now take numbers starting with $78116$ , and write each of them in the form $2^x 5^y r$ . We are looking for 10 consecutive rows where the sum of $y$ s is at least equal to the sum of $x$ s.

 number   x  y  z
 78116    2  0  19529
 78117    0  0  78117
 78118    1  0  39059
 78119:   0  0  78119
 78120:   3  1  1953
 78121:   0  0  78121
 78122:   1  0  39061
 78123:   0  0  78123
 78124:   2  0  19531
 78125:   0  7  1
 78126:   1  0  39063

At this point we can stop, as we just found out that $a_{78117}$ is of the form $2^8 5^8 r_{78117}$ . Therefore the $k$ we seek is $k=78117$ .

Now all we need to do is to compute the last non-zero digit. As the powers of $2$ and $5$ that divide $a_{78117}$ are equal, the last non-zero digit is simply the product of the last digits of the ten $z$ s. This is $7\cdot 9\cdot 9\cdot 3\cdot 1\cdot 1\cdot 3\cdot 1\cdot 1\cdot 3 \equiv\boxed{9}$ .

@@ Line 39: / Line 39: @@
 At this point we can stop, as we just found out that <math>a_{78117}</math> is of the form <math>2^8 5^8 r_{78117}</math>. Therefore the <math>k</math> we seek is <math>k=78117</math>.
-Now all we need to do is to compute the last non-zero digit. As the powers of <math>2</math> and <math>8</math> that divide <math>a_k</math> are equal, the last non-zero digit is simply the product of the last digits of the ten <math>z</math>s. This is <math>7\cdot 9\cdot 9\cdot 3\cdot 1\cdot 1\cdot 3\cdot 1\cdot 1\cdot 3 \equiv\boxed{9}</math>.
+Now all we need to do is to compute the last non-zero digit. As the powers of <math>2</math> and <math>5</math> that divide <math>a_{78117}</math> are equal, the last non-zero digit is simply the product of the last digits of the ten <math>z</math>s. This is <math>7\cdot 9\cdot 9\cdot 3\cdot 1\cdot 1\cdot 3\cdot 1\cdot 1\cdot 3 \equiv\boxed{9}</math>.
 == See also ==
 {{AHSME box|year=1998|num-b=29|after=Last Question}}
 {{MAA Notice}}

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Difference between revisions of "1998 AHSME Problems/Problem 30"

Latest revision as of 09:37, 11 August 2020

Problem

Solution

See also