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Numbers - Modular Arithmetic

For COMPETITION
Number of Total Problems: 12.
FOR PRINT ::: (Book)

Problem Num : 11

From : AMC10

Type:

Section:Numbers

Theme:
Adjustment# : 0

Difficulty: 1

'
The number obtained from the last two nonzero digits of $90!$ is equal to $n$ . What is $n$ ?
$extbf{(A)} 12 qquad extbf{(B)} 32 qquad extbf{(C)} 48 qquad extbf{(D)} 52 qquad extbf{(E)} 68$
'

Category Modular Arithmetic

Analysis

Solution/Answer
We will use the fact that for any integer $n$ , $egin{align*}(5n+1)(5n+2)(5n+3)(5n+4)&=[(5n+4)(5n+1)][(5n+2)(5n+3)]\ &=(25n^2+25n+4)(25n^2+25n+6)equiv 4cdot 6\ ...$
First, we find that the number of factors of $10$ in $90!$ is equal to $leftlfloor frac{90}5 ight floor+leftlfloorfrac{90}{25} ight floor=18+3=21$ . Let $N=frac{90!}{10^{21}}$ . The $n$ we want is therefore the last two digits of $N$ , or $Npmod{100}$ . Since there is clearly an excess of factors of 2, we know that $Nequiv 0pmod 4$ , so it remains to find $Npmod{25}$ .
If we divide $N$ by $5^{21}$ , we can write $N$ as $frac M{2^{21}}$ where $M=1cdot 2cdot 3cdot 4cdot 1cdot 6cdot 7cdot 8cdot 9cdot 2cdots 89cdot 18,$ where every number in the form $(5^a)*n$ is replaced by $n$ .
The number $M$ can be grouped as follows:
$egin{align*}M= &(1cdot 2cdot 3cdot 4)(6cdot 7cdot 8cdot 9)cdots(86cdot 87cdot 88cdot 89)\ &cdot (1cdot...$
Hence, we can reduce $M$ to
$egin{align*}M&equiv(-1)^{18} cdot (-1)^3(16cdot 17cdot 18) cdot (1cdot 2cdot 3) \ &= 1cdot -21cdot 6\ &a...$
Using the fact that $2^{10}=1024equiv -1pmod{25}$ ,we can deduce that $2^{21}equiv 2pmod{25}$ . Therefore $N=frac M{2^{21}}equiv frac {24}2pmod{25}=12pmod{25}$ .
Finally, combining with the fact that $Nequiv 0pmod 4$ yields $n=oxed{ extbf{(A)} 12}$ .

Answer:

Problem Num : 12

From : AMC10

Type:

Section:Numbers

Theme:
Adjustment# : 0

Difficulty: 1

'
A year is a leap year if and only if the year number is divisible by 400 (such as 2000) or is divisible by 4 but not 100 (such as 2012). The 200th anniversary of the birth of novelist Charles Dickens was celebrated on February 7, 2012, a Tuesday. On what day of the week was Dickens born?
$extbf{(A)} ext{Friday}qquad extbf{(B)} ext{Saturday}qquad extbf{(C)} ext{Sunday}qquad extbf{(D)} ext{Mond...$
'

Category Modular Arithmetic

Analysis

Solution/Answer
Each year we go back is one day back, because $365 = 1 ( ext{mod} 7)$ . Each leap year we go back is two days back, since $366 = 2 ( ext{mod} 7)$ . A leap year is GENERALLY every four years, so 200 years would have $frac{200}{4}$ = $50$ leap years, but the problem points out that 1900 does not count as a leap year.
This would mean a total of 151 regular years and 49 leap years, so $1(151)+2(49)$ = $249$ days back. Since $249 = 4 ( ext{mod} 7)$ , four days back from Tuesday would be $oxed{ extbf{(A)} ext{Friday}}$

Answer:

Array ( [0] => 7920 [1] => 7930 ) 121 2