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What is a number that the digits in the number are the same but not 2 digit number nor 3 digit number?
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∙ 9y ago
3333
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∙ 9y ago
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One of the digits in this 2-digit number is 5 but the number is not divisible by 5 nor is it divisible by 9?
51, 52, 53, 56, 57, 58, 59
What is pi to the last 14 digits?
pi is a transcendental number, which is a kind of irrational number. That means that the decimal representation of pi does not end (nor does it have a recurring sequence). There is, therefore, no last digit.
Which 1-digit number is neither a prime nor a square nor a cube?
6
Jenny's number has three digits One digit is a prime number Another digit is a square number The other is digit is neither prime or square Her number is NOT divisable by three What is the grerate?
Single digit primes are 2, 3, 5, 7 Single digit squares are 1, 4, 9 Neither prime nor square are 0, 6, 8 None of that makes a difference if we can't figure out what a "grerate" is. I'll guess you want to know the greatest possible number given those conditions. That would be 985.
What is the largest five-digit number that can be formed with no odd digits and without repeating any digit?
The key to solving this kind of problem is to use the largest digit you possibly can, starting at the left. For example, the largest even digit is 8, so you start with an 8. For the next digit - the second from left - you can't repeat the 8, nor any odd digit, so you need to use the next-largest even digit.
What is the answer to 27-2 when you convert it into a decimal?
29
Is 21 divisible by 2 3 or 5?
21 is not divisible by 2 nor 5, but it is divisible by 3. To be divisible by 2 the last digit must be even (one of {0, 2, 4, 6, 8}; the last digit 1 is not one of these, thus 21 is not divisible by 2 To be divisible by 3 sum the digits and if this total is divisible by 3, then so is the original number; the test can be repeated on the sum, so by repeatedly summing the digits of the totals until a single digit remains only if that single digit is 3, 6 or 9 is the original number divisible by 3. 21 → 2 + 1 = 3 which is divisible by 3 so 21 is divisible by 3. To be divisible by 5, the last digit must be 0 or 5. The last digit of 21 is 1 which is not 0 nor 5, thus 21 is not divisible by 5.
Is 9.565565556 rational?
9.565565556 = 9565565556/1000000000 = 2391391389/250000000 So it is rational. However, if you mean 9.565565556... where it continues with one more 5 than last time followed by a 6 and so on forever, then no, it is not a rational. ----------------------------------------------------------------------------------------- The decimal form of a rational number either terminates or continues with the same one or more digits repeating, eg 1.5, 1.333..., 1.1818181..., 1.1666..., 1.1565656... are all rational numbers. If the decimal does not terminate nor continue forever with the same repeating digit(s) then the number is irrational. 9.565565556 as written terminates and so it a rational number. 9.565565556... does not terminate nor does it continue with the same repeating digits (as an extra 5 is inserted before the next 6), so it it irrational.
How many significant digits are there in 4000000?
There is one significant figure (which I assume you are referring to).However there are 7 digits involved, of which all are significant. Each digit is important and special in its own right. None should be singled out as being different, as that is Digitist.* * * * *Leaving aside the political correctness of the anti-digitism, the number of significant digits depends on the context. In the above example, if it is known that the number is not 3,999,999 nor 4,000,001 then all seven digits are significant. If it is known that the number is 4,000 thousand (not 3,999 thousand or 4,001 thousand) then there are 4 sig digs.
Does 6 go evenly into 35?
No. To be divisible by 6 the number must be divisible by both 2 and 3. To be divisible by 2, the last digit of the number must be even (ie one of {0, 2, 4, 6, 8}). The last digit of 35 is 5 which is not even and so 35 is not divisible by 2. To be divisible by 3, sum the digits of the number and if this sum is divisible by 3 then so is the original number. As the test can be applied to the sum, repeatedly summing the digits of the sums until a single digit remains, then the original number is divisible by 3 only if this single digit is one of {3, 6, 9}. For 35 3 + 5 = 8 which is not divisible by 3 (nor is is one of {3, 6, 9}, thus 35 is not divisible by 3. As 35 is not divisible by both 2 and 3 (in fact it is divisible by neither 2 nor 3) it is not divisible by 6.
Is the number 26 divisible by 2 3 5 or 9 How do you know?
To be divisible by 2 the number must be even, that is its last digit must be 2, 4, 6, 8, or 0; the last digit of 26 is 6 which is one of {2, 4, 6, 8, 0} so 26 is even and divisible by 2. To be divisible by 3 sum the digits of the number; if this sum is divisible by 3 then the original number is divisible by 3. The test can be repeated on the sum until a single digit remains, in which case if this single digit is 3, 6, or 9 then the original number is divisible by 3; For 26: 2 + 6 = 8 which is not one of {3, 6, 9} so 26 is not divisible by 3. To be divisible by 5 the last digit must be a 0 or 5; the last digit of 26 is a 6 which is not 0 nor 5, so 26 is not divisible by 5. To be divisible by 9 sum the digits of the number; if this sum is divisible by 9 then the original number is divisible by 9. The test can be repeated on the sum until a single digit remains, in which case if this single digit is 9 then the original number is divisible by 9; For 26: 2 + 6 = 8 which is not 9 so 26 is not divisible by 9. 26 is divisible by 2 but not divisible by 3, 5 nor 9.
Is 49 an integer?
Basically, an integer is a number that does NOT have digits after the decimal point (nor should it have a fractional part).
