The multiples of 15 are the numbers which are n-times of 15, where n = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 and so on. So, basically, multiples are n-times of any number, where n is the list of natural numbers. Thus, the number 15 has multiples in the form of 15n, such that: Show
and so on. Here is the list of all multiples:
All numbers can be divided or the product of 15 is a multiple of 15. The multiples are different from the factors of 15. The factors are the numbers which when multiplied in a pair give the original number such as 3 x 5 = 15, so here 3 and 5 are not the multiples but the factors. Multiples do not produce the original number instead they are divisible from the original number, for example, 45 ÷ 15 = 3. What is a multiple of 15?Any number that can be denoted in the form 15n where n is an integer is a multiple of 15. So if two values p and q, we say that q is a multiple of p if q = np for some integer n. For example, 60, 45, 150 and 120 are all multiples of 15 for the following reasons
These values are called the multiples as these values are obtained by adding or subtracting the original value many times. What is the fifth multiple of 15?The fifth multiple of 15 could be found by multiplying 15 by 5, such that; 15 x 5 = 75 Thus, the fifth multiple of 15 is 75. Multiples of 15 ChartHere is the table of all the multiples of integer 15 from 1 to 20. Students can try writing the multiples till 100 by just multiplying 15 with natural numbers up to 100 to have practice.
Facts about Multiples of 15Some interesting facts about multiples of 15 are given below:
Also, check the multiples of some numbers given here.
Video Lesson on Common Multiples Solved Examples on Multiples of 15Question 1: Dena visits the pool for swimming once every thirty days and Rachael once in fifteen days. If we assume that they go swimming on the first day of May, when will their next possible meet together at the pool? Solution: The common multiple of the numbers 15 and 30 are given below. The multiples of 15 are 15, 30, 45,….. The multiples of 30 are 30, 60, 90,….. The multiples in common to 30 and 15 are 30, 60, 90, ….. The smallest common multiple of the numbers 30 and 15 is 30. They visited the pool and swam together on the first day of May. So, their next possible meet is 1 + 30 = 31 May Question 2: Reena can eat 15 candies in a day. Find the maximum number of candies that she could eat in 14 days. Solution: Number of candies that Reena can eat in a day = 15 The maximum number of candies that she could eat in 14 days = 14th multiple of 15 = 15 × 14 = 210 Register at BYJU’S to learn more on multiples of other numbers. The first five multiples of 15 are 15, 30, 45, 60 and 75. These can be obtained as: 15 × 1 = 15 15 × 2 = 30 15 × 3 = 45 15 × 4 = 60 15 × 5 = 75 The 8th multiple of 15 can be written as 15 × 8 = 120. We can find the 20th multiple of 15 by multiplying 15 by 20. This can be done as: 15 × 20 = 300 Thus, the 20th multiple of 15 is 300. A triangular number or triangle number counts objects arranged in an equilateral triangle. Triangular numbers are a type of figurate number, other examples being square numbers and cube numbers. The nth triangular number is the number of dots in the triangular arrangement with n dots on each side, and is equal to the sum of the n natural numbers from 1 to n. The sequence of triangular numbers, starting with the 0th triangular number, is
0, 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120, 136, 153, 171, 190, 210, 231, 253, 276, 300, 325, 351, 378, 406, 435, 465, 496, 528, 561, 595, 630, 666... (This sequence is included in the On-Line Encyclopedia of Integer Sequences (sequence A000217 in the OEIS).) Derivation of triangular numbers from a left-justified Pascal's triangle The triangular numbers are given by the following explicit formulas: T n = ∑ k = 1 n k = 1 + 2 + 3 + ⋯ + n = n ( n + 1 ) 2 = ( n + 1 2 ) , {\displaystyle T_{n}=\sum _{k=1}^{n}k=1+2+3+\dotsb +n={\frac {n(n+1)}{2}}={n+1 \choose 2},} where ( n + 1 2 ) {\displaystyle \textstyle {n+1 \choose 2}} is a binomial coefficient. It represents the number of distinct pairs that can be selected from n + 1 objects, and it is read aloud as "n plus one choose two".The first equation can be illustrated using a visual proof.[1] For every triangular number T n {\displaystyle T_{n}} , imagine a "half-square" arrangement of objects corresponding to the triangular number, as in the figure below. Copying this arrangement and rotating it to create a rectangular figure doubles the number of objects, producing a rectangle with dimensions n × ( n + 1 ) {\displaystyle n\times (n+1)} , which is also the number of objects in the rectangle. Clearly, the triangular number itself is always exactly half of the number of objects in such a figure, or: T n = n ( n + 1 ) 2 {\displaystyle T_{n}={\frac {n(n+1)}{2}}} . The example T 4 {\displaystyle T_{4}} follows: 2 T 4 = 4 ( 4 + 1 ) = 20 {\displaystyle 2T_{4}=4(4+1)=20} (green plus yellow) implies that T 4 = 4 ( 4 + 1 ) 2 = 10 {\displaystyle T_{4}={\frac {4(4+1)}{2}}=10} (green). The first equation can also be established using mathematical induction:[2] Since T 1 {\displaystyle T_{1}} is equal to one, a basis case is established. It follows from the definition that T n = n + T n − 1 {\displaystyle T_{n}=n+T_{n-1}} , so assuming the inductive hypothesis for n − 1 {\displaystyle n-1} , adding n {\displaystyle n} to both sides immediately gives T n = n + T n − 1 = 2 n 2 + ( n − 1 ) n 2 = ( 2 + n − 1 ) n 2 = ( n + 1 ) n 2 . {\displaystyle T_{n}=n+T_{n-1}={\frac {2n}{2}}+{\frac {(n-1)n}{2}}={\frac {(2+n-1)n}{2}}={\frac {(n+1)n}{2}}.} In other words, since the proposition P ( n ) {\displaystyle P(n)} (that is, the first equation, or inductive hypothesis itself) is true when n = 1 {\displaystyle n=1} , and since P ( n ) {\displaystyle P(n)} being true implies that P ( n + 1 ) {\displaystyle P(n+1)} is also true, then the first equation is true for all natural numbers. The above argument can be easily modified to start with, and include, zero. The German mathematician and scientist, Carl Friedrich Gauss, is said to have found this relationship in his early youth, by multiplying n/2 pairs of numbers in the sum by the values of each pair n + 1.[3] However, regardless of the truth of this story, Gauss was not the first to discover this formula, and some find it likely that its origin goes back to the Pythagoreans in the 5th century BC.[4] The two formulas were described by the Irish monk Dicuil in about 816 in his Computus.[5] An English translation of Dicuil's account is available.[6] The triangular number Tn solves the handshake problem of counting the number of handshakes if each person in a room with n + 1 people shakes hands once with each person. In other words, the solution to the handshake problem of n people is Tn−1.[7] The function T is the additive analog of the factorial function, which is the products of integers from 1 to n. The number of line segments between closest pairs of dots in the triangle can be represented in terms of the number of dots or with a recurrence relation: L n = 3 T n − 1 = 3 ( n 2 ) ; L n = L n − 1 + 3 ( n − 1 ) , L 1 = 0. {\displaystyle L_{n}=3T_{n-1}=3{n \choose 2};~~~L_{n}=L_{n-1}+3(n-1),~L_{1}=0.} In the limit, the ratio between the two numbers, dots and line segments is lim n → ∞ T n L n = 1 3 . {\displaystyle \lim _{n\to \infty }{\frac {T_{n}}{L_{n}}}={\frac {1}{3}}.} Triangular numbers have a wide variety of relations to other figurate numbers. Most simply, the sum of two consecutive triangular numbers is a square number, with the sum being the square of the difference between the two (and thus the difference of the two being the square root of the sum). Algebraically, T n + T n − 1 = ( n 2 2 + n 2 ) + ( ( n − 1 ) 2 2 + n − 1 2 ) = ( n 2 2 + n 2 ) + ( n 2 2 − n 2 ) = n 2 = ( T n − T n − 1 ) 2 . {\displaystyle T_{n}+T_{n-1}=\left({\frac {n^{2}}{2}}+{\frac {n}{2}}\right)+\left({\frac {\left(n-1\right)^{2}}{2}}+{\frac {n-1}{2}}\right)=\left({\frac {n^{2}}{2}}+{\frac {n}{2}}\right)+\left({\frac {n^{2}}{2}}-{\frac {n}{2}}\right)=n^{2}=(T_{n}-T_{n-1})^{2}.}This fact can be demonstrated graphically by positioning the triangles in opposite directions to create a square: 6 + 10 = 16 10 + 15 = 25 The double of a triangular number, as in the visual proof from the above section § Formula, is called a pronic number. There are infinitely many triangular numbers that are also square numbers; e.g., 1, 36, 1225. Some of them can be generated by a simple recursive formula: S n + 1 = 4 S n ( 8 S n + 1 ) {\displaystyle S_{n+1}=4S_{n}\left(8S_{n}+1\right)} with S 1 = 1. {\displaystyle S_{1}=1.}All square triangular numbers are found from the recursion S n = 34 S n − 1 − S n − 2 + 2 {\displaystyle S_{n}=34S_{n-1}-S_{n-2}+2} with S 0 = 0 {\displaystyle S_{0}=0} and S 1 = 1. {\displaystyle S_{1}=1.}A square whose side length is a triangular number can be partitioned into squares and half-squares whose areas add to cubes. This shows that the square of the nth triangular number is equal to the sum of the first n cube numbers. Also, the square of the nth triangular number is the same as the sum of the cubes of the integers 1 to n. This can also be expressed as ∑ k = 1 n k 3 = ( ∑ k = 1 n k ) 2 . {\displaystyle \sum _{k=1}^{n}k^{3}=\left(\sum _{k=1}^{n}k\right)^{2}.}The sum of the first n triangular numbers is the nth tetrahedral number: ∑ k = 1 n T k = ∑ k = 1 n k ( k + 1 ) 2 = n ( n + 1 ) ( n + 2 ) 6 . {\displaystyle \sum _{k=1}^{n}T_{k}=\sum _{k=1}^{n}{\frac {k(k+1)}{2}}={\frac {n(n+1)(n+2)}{6}}.} More generally, the difference between the nth m-gonal number and the nth (m + 1)-gonal number is the (n − 1)th triangular number. For example, the sixth heptagonal number (81) minus the sixth hexagonal number (66) equals the fifth triangular number, 15. Every other triangular number is a hexagonal number. Knowing the triangular numbers, one can reckon any centered polygonal number; the nth centered k-gonal number is obtained by the formula C k n = k T n − 1 + 1 {\displaystyle Ck_{n}=kT_{n-1}+1} where T is a triangular number. The positive difference of two triangular numbers is a trapezoidal number. The pattern found for triangular numbers ∑ n 1 = 1 n 2 n 1 = ( n 2 + 1 2 ) {\displaystyle \sum _{n_{1}=1}^{n_{2}}n_{1}={\binom {n_{2}+1}{2}}} and for tetrahedral numbers ∑ n 2 = 1 n 3 ∑ n 1 = 1 n 2 n 1 = ( n 3 + 2 3 ) , {\displaystyle \sum _{n_{2}=1}^{n_{3}}\sum _{n_{1}=1}^{n_{2}}n_{1}={\binom {n_{3}+2}{3}},} which uses binomial coefficients, can be generalized. This leads to the formula:[8] ∑ n k − 1 = 1 n k ∑ n k − 2 = 1 n k − 1 … ∑ n 2 = 1 n 3 ∑ n 1 = 1 n 2 n 1 = ( n k + k − 1 k ) {\displaystyle \sum _{n_{k-1}=1}^{n_{k}}\sum _{n_{k-2}=1}^{n_{k-1}}\ldots \sum _{n_{2}=1}^{n_{3}}\sum _{n_{1}=1}^{n_{2}}n_{1}={\binom {n_{k}+k-1}{k}}} Triangular numbers correspond to the first-degree case of Faulhaber's formula. Alternating triangular numbers (1, 6, 15, 28, ...) are also hexagonal numbers. Every even perfect number is triangular (as well as hexagonal), given by the formula M p 2 p − 1 = M p ( M p + 1 ) 2 = T M p {\displaystyle M_{p}2^{p-1}={\frac {M_{p}(M_{p}+1)}{2}}=T_{M_{p}}} where Mp is a Mersenne prime. No odd perfect numbers are known; hence, all known perfect numbers are triangular.For example, the third triangular number is (3 × 2 =) 6, the seventh is (7 × 4 =) 28, the 31st is (31 × 16 =) 496, and the 127th is (127 × 64 =) 8128. The final digit of a triangular number is 0, 1, 3, 5, 6, or 8, and thus never end in 2, 4, 7, or 9. A final 3 must be preceded by a 0 or 5; a final 8 must be preceded by a 2 or 7. In base 10, the digital root of a nonzero triangular number is always 1, 3, 6, or 9. Hence, every triangular number is either divisible by three or has a remainder of 1 when divided by 9: 0 = 9 × 0 1 = 9 × 0 + 1 3 = 9 × 0 + 3 6 = 9 × 0 + 6 10 = 9 × 1 + 1 15 = 9 × 1 + 6 21 = 9 × 2 + 3 28 = 9 × 3 + 1 36 = 9 × 4 45 = 9 × 5 55 = 9 × 6 + 1 66 = 9 × 7 + 3 78 = 9 × 8 + 6 91 = 9 × 10 + 1 ...There is a more specific property to the triangular numbers that aren't divisible by 3; that is, they either have a remainder 1 or 10 when divided by 27. Those that are equal to 10 mod 27 are also equal to 10 mod 81. The digital root pattern for triangular numbers, repeating every nine terms, as shown above, is "1, 3, 6, 1, 6, 3, 1, 9, 9". The converse of the statement above is, however, not always true. For example, the digital root of 12, which is not a triangular number, is 3 and divisible by three. If x is a triangular number, then ax + b is also a triangular number, given a is an odd square and b = a − 1/8. Note that b will always be a triangular number, because 8Tn + 1 = (2n + 1)2, which yields all the odd squares are revealed by multiplying a triangular number by 8 and adding 1, and the process for b given a is an odd square is the inverse of this operation. The first several pairs of this form (not counting 1x + 0) are: 9x + 1, 25x + 3, 49x + 6, 81x + 10, 121x + 15, 169x + 21, ... etc. Given x is equal to Tn, these formulas yield T3n + 1, T5n + 2, T7n + 3, T9n + 4, and so on. The sum of the reciprocals of all the nonzero triangular numbers is ∑ n = 1 ∞ 1 n 2 + n 2 = 2 ∑ n = 1 ∞ 1 n 2 + n = 2. {\displaystyle \sum _{n=1}^{\infty }{1 \over {{n^{2}+n} \over 2}}=2\sum _{n=1}^{\infty }{1 \over {n^{2}+n}}=2.} This can be shown by using the basic sum of a telescoping series: ∑ n = 1 ∞ 1 n ( n + 1 ) = 1. {\displaystyle \sum _{n=1}^{\infty }{1 \over {n(n+1)}}=1.} Two other formulas regarding triangular numbers are T a + b = T a + T b + a b {\displaystyle T_{a+b}=T_{a}+T_{b}+ab} andT a b = T a T b + T a − 1 T b − 1 , {\displaystyle T_{ab}=T_{a}T_{b}+T_{a-1}T_{b-1},} both of which can easily be established either by looking at dot patterns (see above) or with some simple algebra.In 1796, Gauss discovered that every positive integer is representable as a sum of three triangular numbers (possibly including T0 = 0), writing in his diary his famous words, "ΕΥΡΗΚΑ! num = Δ + Δ + Δ". This theorem does not imply that the triangular numbers are different (as in the case of 20 = 10 + 10 + 0), nor that a solution with exactly three nonzero triangular numbers must exist. This is a special case of the Fermat polygonal number theorem. The largest triangular number of the form 2k − 1 is 4095 (see Ramanujan–Nagell equation). Wacław Franciszek Sierpiński posed the question as to the existence of four distinct triangular numbers in geometric progression. It was conjectured by Polish mathematician Kazimierz Szymiczek to be impossible and was later proven by Fang and Chen in 2007.[9][10] Formulas involving expressing an integer as the sum of triangular numbers are connected to theta functions, in particular the Ramanujan theta function.[11][12] The maximum number of pieces, p obtainable with n straight cuts is the n-th triangular number plus one, forming the lazy caterer's sequence (OEIS A000124) A fully connected network of n computing devices requires the presence of Tn − 1 cables or other connections; this is equivalent to the handshake problem mentioned above. In a tournament format that uses a round-robin group stage, the number of matches that need to be played between n teams is equal to the triangular number Tn − 1. For example, a group stage with 4 teams requires 6 matches, and a group stage with 8 teams requires 28 matches. This is also equivalent to the handshake problem and fully connected network problems. One way of calculating the depreciation of an asset is the sum-of-years' digits method, which involves finding Tn, where n is the length in years of the asset's useful life. Each year, the item loses (b − s) × n − y/Tn, where b is the item's beginning value (in units of currency), s is its final salvage value, n is the total number of years the item is usable, and y the current year in the depreciation schedule. Under this method, an item with a usable life of n = 4 years would lose 4/10 of its "losable" value in the first year, 3/10 in the second, 2/10 in the third, and 1/10 in the fourth, accumulating a total depreciation of 10/10 (the whole) of the losable value. By analogy with the square root of x, one can define the (positive) triangular root of x as the number n such that Tn = x:[13] n = 8 x + 1 − 1 2 {\displaystyle n={\frac {{\sqrt {8x+1}}-1}{2}}} which follows immediately from the quadratic formula. So an integer x is triangular if and only if 8x + 1 is a square. Equivalently, if the positive triangular root n of x is an integer, then x is the nth triangular number.[13] An alternative name proposed by Donald Knuth, by analogy to factorials, is "termial", with the notation n? for the nth triangular number.[14] However, although some other sources use this name and notation,[15] they are not in wide use.
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Two natural numbers whose difference is 25 and the least common multiple is 630 find the numbers
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