-What I know: A nonzero integer d is said to divide an integer a (d|a) if there
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Question
-What I know: A nonzero integer d is said to divide an integer a (d|a) if there exists an integer b such that a=db. If d divides a, then d is referred to as a divisor of a or a factor of a, and a is referred to as a multiple of d
-If d is a nonzero integer such that d|a and d}b for two integers a and b, then for any integers x and y, d}(xa +yb) (in particular, d|(a+b) and d|(a-b)
-Division algorithim
-definition of gcd
Prove the following:
1 Prove that an integer is divisible by 4 IFF the integer represented by the tens digit and the units digit is divisible by 4. (To give you an example, the "integer represented by the tens digit and the units digit" of 1024 is 24, and the assertion is that 1024 is divisible by 4 IFF 24 is divisible by 4- which it is!)
2 Prove that an integer is divisible by 8 IFF the integer represented by the thousands digit and the tens digit and the units digit is divisible by 8.
3 Prove that an integer is divisible by 3 IFF the sum of its digits is divisible by 3. (ie sum of digits of 1024 is 1 +0 + 2+ 4 = 7 and the assertion is that 1024 is divisible by 3 IFF 7 is divisible by 3- and therefore, since 7 is not divisible by 3 we can conclude that 1024 is not divisible by 3 either! Hint: in context of this problem: 1024 = 1 x 1000 + 0 x 100 + 2 x 10 +4= 1 (999 +1) + 0x (99 +1 ) + 2(9+1) +4. What can you say about the terms containing 9 ,99, 999 as far as divisibility by 3 is concerned. (Use: if d is a nonzero integer such that d|a and d|b for two integers a and b, then for any integers x and y, d|(xa+ yb). (in particular, d|(a+b) and d|(a-b). ))
4 Prove that an integer is divisible by 9 IFF the sum of its digits is divisible by 9
5 Prove that an integer is divisible by 11 IFF the difference between the sum of the digits in the units place, the hundreds place, the ten thousands place, ...( the places corresponding to the even powers of 10 ) and the sum of the digits in the tens place, the thousands place, the hundred thousands place, ... (the places corresponding to the odd powers of 10 ) is divisible by 11 (hint 10= 11-1, 100= 99+1, 1000= 1001-1, 10000=9999+1, etc. What can you say about the integers 11, 99, 1001, 9999, etc as far as divisibility is concerned?)
Explanation / Answer
-What I know: A nonzero integer d is said to divide an integer a (d|a) if there exists an integer b such that a=db. If d divides a, then d is referred to as a divisor of a or a factor of a, and a is referred to as a multiple of d
-If d is a nonzero integer such that d|a and d}b for two integers a and b, then for any integers x and y, d}(xa +yb) (in particular, d|(a+b) and d|(a-b)
-Division algorithim
-definition of gcd
Prove the following:
1 Prove that an integer is divisible by 4 IFF the integer represented by the tens digit and the units digit is divisible by 4. (To give you an example, the "integer represented by the tens digit and the units digit" of 1024 is 24, and the assertion is that 1024 is divisible by 4 IFF 24 is divisible by 4- which it is!)
LET THE LAST 2 DIGITS OF ANY NUMBER BE XY....ITS VALUE = 10X+Y
LET THE NUMBER BE
N= {P*******Q}XY
WHERE THERE COULD BE ANY NUMBER OF * S AND THEY COULD BE ANY NUMBERS .IN GENERAL [P*******Q] IS AN INTEGER >=0
SO ITS VALUE
N= [100*AN INTEGER GIVEN BY {P*******Q}]+10X+Y
PROPOSITION
10X+Y IS DIVISIBLE BY 4 THEN WE GET
10X+Y=4A WHERE A IS AN INTEGER[100*AN INTEGER GIVEN BY {P*******Q}]
OBVIOUSLY 100*AN INTEGER GIVEN BY [{P*******Q}] IS DIVISIBLE BY 4 SINCE 4|100
SO [{P*********Q}]=4B WHERE B IS AN INTEGER...SO
N=4B+4A=4[A+B]=4C WHERE C=A+B IS AN INTEGER...SO
4 | N ...................PROVED
CONVERSE
4|N.....N=4*C WHERE C IS AN INTEGER
AS PROVED ABOVE, WE HAVE
N=4C=4B+[10X+Y]
10X+Y=4[C-B]=4*A WHERE A=C-B IS AN INTEGER....SO..
4|10X+Y....PROVED
2.Prove that an integer is divisible by 8 IFF the integer represented by the thousands digit and the tens digit and the units digit is divisible by 8.
PROOF IS SIMILAR TO THAT ABOVE
TAKE LAST 3 DIGITS AS XYZ....VALUE=100X+10Y+Z......
NUMBER = N = [{P**********Q}]XYZ....ITS VALUE IS
N= 1000*[{P**********Q}]+100X+10Y+Z....ETC....
3 Prove that an integer is divisible by 3 IFF the sum of its digits is divisible by 3. (ie sum of digits of 1024 is 1 +0 + 2+ 4 = 7 and the assertion is that 1024 is divisible by 3 IFF 7 is divisible by 3- and therefore, since 7 is not divisible by 3 we can conclude that 1024 is not divisible by 3 either! Hint: in context of this problem: 1024 = 1 x 1000 + 0 x 100 + 2 x 10 +4= 1 (999 +1) + 0x (99 +1 ) + 2(9+1) +4. What can you say about the terms containing 9 ,99, 999 as far as divisibility by 3 is concerned. (Use: if d is a nonzero integer such that d|a and d|b for two integers a and b, then for any integers x and y, d|(xa+ yb). (in particular, d|(a+b) and d|(a-b). ))
LET ANY NUMBER BE ..WE ARE TAKING BELOW A 5 DIGIT NUMBER FOR A CONCRETE EXAMPLE ..BUT THE PROOF IS SAME FOR ANY NUMBER OF DIGITS
LET THE NUMBER N BE PQRST...SO VALUE OF ..
N= 1*T+10*S+100*R+1000*Q+10000*P = 1*T+[9+1]S+[99+1]R+[999+1]Q+[9999+1]P
N= [P+Q+R+S+T]+9[S+11R+111Q+1111P]=[P+Q+R+S+T]+9*A, WHERE
A=S+11R+111Q+1111P IS AN INTEGER
PROPOSITION
P+Q+R+S+T=SUM OF DIGITS IS DIVISIBLE BY 3
P+Q+R+S+T=3B WHERE A IS AN INTEGER...SO
N=3B+9A=3[B+3A]=3C, WHERE C=B+3A IS AN INTEGER....SO
3|N ...................PROVED
CONVERSE
3|N ...LET N=3C
SO WE HAVE
N=3C=[P+Q+R+S+T]+9A
[P+Q+R+S+T]=3[C-3A]=3B WHERE B = C-3A IS AN INTEGER...SO
3|[P+Q+R+S+T]......PROVED
4 Prove that an integer is divisible by 9 IFF the sum of its digits is divisible by 9
PROOF IS SAME AS ABOVE EXCEPT THAT WE WRITE
P+Q+R+S+T = 9B...ETC...
5 Prove that an integer is divisible by 11 IFF the difference between the sum of the digits in the units place, the hundreds place, the ten thousands place, ...( the places corresponding to the even powers of 10 ) and the sum of the digits in the tens place, the thousands place, the hundred thousands place, ... (the places corresponding to the odd powers of 10 ) is divisible by 11 (hint 10= 11-1, 100= 99+1, 1000= 1001-1, 10000=9999+1, etc. What can you say about the integers 11, 99, 1001, 9999, etc as far as divisibility is concerned?)
LET ANY NUMBER BE ..WE ARE TAKING BELOW A 5 DIGIT NUMBER FOR A CONCRETE EXAMPLE ..BUT THE PROOF IS SAME FOR ANY NUMBER OF DIGITS
LET THE NUMBER N BE PQRST...SO VALUE OF ..
N= 1*T+10*S+100*R+1000*Q+10000*P = 1*T+[11-1]S+[99+1]R+[1001-1]Q+[9999+1]P
N= [P-Q+R-S+T]+11[S+9R+91Q+909P]=[P-Q+R-S+T]+11*A, WHERE
A=S-9R+91Q+909P IS AN INTEGER
PROPOSITION
P-Q+R-S+T=11B ...AS GIVEN ........SO
N=11B+11A=11[B+A]=11C, WHERE C=B+A IS AN INTEGER....SO
11|N ...................PROVED
CONVERSE
11|N ...LET N=11C
SO WE HAVE
N=11C=[P-Q+R-S+T]+11A
[P-Q+R-S+T]=11[C-A]=11B WHERE B = C-A IS AN INTEGER...SO
11|[P-Q+R-S+T]......PROVED
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