Question 1210170: How many zeroes do we write when we write all the integers from 1 to 625 in base 4?
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let's solve this problem by analyzing the number of zeroes in each place value when writing the numbers from 1 to 625 in base 4.
**1. Convert 625 to Base 4**
* $625 \div 4 = 156$ remainder 1
* $156 \div 4 = 39$ remainder 0
* $39 \div 4 = 9$ remainder 3
* $9 \div 4 = 2$ remainder 1
* $2 \div 4 = 0$ remainder 2
So, $625_{10} = 21301_4$. This means the largest base 4 number we're dealing with has 5 digits.
**2. Analyze Place Values**
We need to count the number of zeroes in each place value (units, fours, sixteens, sixty-fours, and two-hundred-fifty-sixes).
* **Units Place (4^0):**
* Numbers ending in 0 in base 4 are multiples of 4.
* From 1 to 625, there are $\lfloor 625/4 \rfloor = 156$ multiples of 4. So, there are 156 zeroes in the units place.
* **Fours Place (4^1):**
* We are looking for numbers with a 0 in the fours place.
* We look at the pattern in base 4: \_ \_ 0 \_
* For every 16 numbers, there are 4 numbers with 0 in the fours place.
* From 1 to 624, we have $624/16 = 39$ sets of 16. In each set, there are 4 zeroes. So, $39 \times 4 = 156$ zeroes.
* 625 is $21301_4$ so it also includes one zero in the fours place.
* total 156+1=157 zeroes
* **Sixteens Place (4^2):**
* We are looking for numbers with a 0 in the sixteens place.
* We look at the pattern in base 4: \_ 0 \_ \_
* For every 64 numbers, there are 16 numbers with 0 in the sixteens place.
* From 1 to 576 we have $576/64 = 9$ sets of 64. In each set there are 16 zeroes.
* $9 \times 16 = 144$ zeroes.
* From 577 to 625, the numbers in base 4 are:
* 21001, 21002, 21003, 21010, 21011, 21012, 21013, 21020, 21021, 21022, 21023, 21030, 21031, 21032, 21033, 21100, 21101, 21102, 21103, 21110, 21111, 21112, 21113, 21120, 21121, 21122, 21123, 21130, 21131, 21132, 21133, 21200, 21201, 21202, 21203, 21210, 21211, 21212, 21213, 21220, 21221, 21222, 21223, 21230, 21231, 21232, 21233, 21300, 21301.
* There are 13 more numbers with a zero in the sixteens place.
* 144+13 = 157 zeroes.
* **Sixty-fours Place (4^3):**
* We are looking for numbers with a 0 in the sixty-fours place.
* We look at the pattern in base 4: 0 \_ \_ \_
* For every 256 numbers, there are 64 numbers with 0 in the sixty-fours place.
* From 1 to 512, there are $512/256 = 2$ sets of 256. In each set, there are 64 zeroes.
* $2 \times 64 = 128$ zeroes.
* From 513 to 625, we have
* 20001 to 21301.
* 20001, 20002, 20003, 20010, 20011, 20012, 20013, 20020, 20021, 20022, 20023, 20030, 20031, 20032, 20033, 20100, 20101, 20102, 20103, 20110, 20111, 20112, 20113, 20120, 20121, 20122, 20123, 20130, 20131, 20132, 20133, 20200, 20201, 20202, 20203, 20210, 20211, 20212, 20213, 20220, 20221, 20222, 20223, 20230, 20231, 20232, 20233, 21000, 21001, 21002, 21003, 21010, 21011, 21012, 21013, 21020, 21021, 21022, 21023, 21030, 21031, 21032, 21033.
* We have 49 numbers with a zero in the 64's place.
* 128+49=177 zeroes
* **Two-hundred-fifty-sixes Place (4^4):**
* We are looking for numbers with a 0 in the 256's place.
* We look at the pattern in base 4: 0 \_ \_ \_ \_
* From 1 to 255, there are 255 numbers and no numbers with a zero in the 4^4 position.
* From 256 to 511 we have 10000 to 13333, and no zeros in the 256 position.
* From 512 to 625, we have 20000 to 21301. all of these numbers have a 2 in the 256's place, and no zeroes.
* 0 zeroes.
**3. Total Zeroes**
* Total zeroes = 156 + 157 + 157 + 177 + 0 = 647.
**Therefore, we write 647 zeroes when writing all the integers from 1 to 625 in base 4.**
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