The 2-digit number must be 20, because it is the only 2-digit number whose sum of its two even digits, 2 + 0 = 2, is greater than the product of its two even digits, 2 x 0 = 0. Moreover, 20 is a product of the two consecutive integers 4 and 5.
-23
0 and 9 or, 0 and anything really.
Yes: 1 and 0 are consecutive.
420
To find the number of 7-digit positive integers where the product of the digits equals 10,000, we start by factoring 10,000 into its prime factors: (10,000 = 10^4 = (2 \cdot 5)^4 = 2^4 \cdot 5^4). The digits can only be from 1 to 9, so we can use digits like 1, 2, 4, 5, 8, etc., that can contribute to this product. Combinations of these digits that yield a product of 10,000 must be examined, but the key constraint is that the total number of digits must equal 7. This requires a systematic approach to explore all valid combinations of digits that meet these criteria, which can be complex and requires further analysis involving combinatorial counting and digital constraints. The exact count of such integers is not straightforward without deeper calculations and could vary significantly based on the digit choices made.
Integers of 6 digits are normally greater than integers of 5 digits
Three of them.
Find the greatest product of five consecutive digits in the 1000-digit number.7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450
There is no pair of consecutive even integers whose sum is 264 - as their average is divisible by 4, that must mean that the sum of the unit digits of each of the pair of numbers could never be a multiple of 4 - which 264 is.
-23
81
I'm assuming the question should read n greater than 10 and less than 100 and there are 8 numbers that satisfy this,1223344556677889So the answer is dhttp://www.webanswers.com/share-question.cfm?q=how-many-integers-n-greater-than-and-less-than-100-are-there-such-that-if-the-digits-of-n-are-the-is-1379ff| http://www.webanswers.com/answer/1566078/education/how-many-integers-n-greater-than-and-less-than-100-are-there-such-that-if-the-digits-of-n-are-the-is-1379ff| http://www.webanswers.com/report-abuse.cfm?q=how-many-integers-n-greater-than-and-less-than-100-are-there-such-that-if-the-digits-of-n-are-the-is-1379ff&p=1566078
43
0 and 9 or, 0 and anything really.
The number is 56
Yes: 1 and 0 are consecutive.
420